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遗传算法旅行商问题c语言代码

遗传算法是一种模拟自然选择过程的优化算法，可以用于解决各种复杂的组合优化问题。

其中，旅行商问题是一个经典的组合优化问题，也是一个典型的NP难题，即寻找最优解的时间复杂度是指数级的。

在本文中，我们将讨论如何使用遗传算法来解决旅行商问题，并给出相应的C语言代码实现。

我们将介绍旅行商问题的数学模型，然后简要介绍遗传算法的原理，最后给出C语言代码实现。

旅行商问题是指一个旅行商要拜访n个城市，恰好拜访每个城市一次，并返回出发城市，要求总路程最短。

数学上可以用一个n*n的距离矩阵d[i][j]表示城市i到城市j的距离，问题可以形式化为求解一个排列p={p1,p2,...,pn}，使得目标函数f(p)=Σd[p[i]][p[i+1]]+d[p[n]][p[1]]最小。

这个问题是一个组合优化问题，其搜索空间是一个n维的离散空间。

遗传算法是一种基于生物进化过程的优化算法，主要包括选择、交叉、变异等操作。

在使用遗传算法解决旅行商问题时，可以将每个排列p看作一个个体，目标函数f(p)看作个体的适应度，通过选择、交叉和变异等操作来搜索最优解。

以下是遗传算法解决旅行商问题的C语言代码实现：1. 我们需要定义城市的距离矩阵和其他相关参数，例如城市的数量n，种裙大小pop_size，交叉概率pc，变异概率pm等。

2. 我们初始化种裙，即随机生成pop_size个排列作为初始种裙。

3. 我们进入遗传算法的迭代过程。

在每一代中，我们首先计算种裙中每个个体的适应度，然后通过选择、交叉和变异操作来更新种裙。

4. 选择操作可以采用轮盘赌选择法，即根据个体的适应度来进行选择，适应度越高的个体被选中的概率越大。

5. 交叉操作可以采用部分映射交叉方法，即随机选择两个个体，然后随机选择一个交叉点，将交叉点之后的基因片段进行交换。

6. 变异操作可以采用变异率为pm的单点变异方法，即随机选择一个个体和一个位置，将该位置的基因值进行随机变异。

7. 我们重复进行迭代操作，直到达到停止条件（例如达到最大迭代次数或者适应度达到阈值）。

编程：Lingo旅行商问题

编程：Lingo旅行商问题MODEL:! 旅行商问题：从城市O出发经过城市A, B, C, D,E,F后又回到O 的过程;SETS:CITY / O A B C D E F/: P;PRED( CITY,CITY)/A,B C,D E,F/旅行优先顺序;STEP/1..7/STEP（I）表示第I个经过的城市;LINK( CITY, CITY):T, ! 时间矩阵;X; ! X( I, J);TXS( CITY,STEP):Y;ENDSETSDATA: !时间矩阵;T = 0 1.7 1.5 1.4 3 1.9 2.81.7 0 32.6 4 33.51.5 3 0 1.5 3.72.6 2.81.42.6 1.5 0 11.5 33.23 4 3.7 11.5 0 3.6 3.31.9 32.6 33.6 0 15.62.83.5 2.8 3.2 3.3 15.6 0;ENDDATA!城市个数;N = @SIZE( CITY);MIN = @SUM( LINK:T * X);!第K个经过城市I;@FOR( CITY( I): @SUM( STEP( K): Y( I, K)) = 1);!城市A和B经过的次序相邻，同理C和D，E和F;@FOR( PRED( I, J): @SUM( STEP( K): P(K) * Y( J, K) - P(K) * Y( I, K)) =1#OR#-1);@FOR( CITY( K):!进入城区;@SUM( CITY( I)| I #NE# K: X( I, K)) = 1;!离开城区;@SUM( CITY( J)| J #NE# K: X( K, J)) = 1;@FOR( CITY( J)| J #GT# 1 #AND# J #NE# K:P( J) >= P( K) + X ( K, J) -( N - 2) * ( 1 - X( K, J)) +( N - 3) * X( J, K)););! 使X、Y取0或1;@FOR( LINK: @BIN( X));@FOR( TXS: @BIN( Y));!出发点和终点;@FOR( CITY( K)| K #GT# 1:P( K) <= N - 1 - ( N - 2) * X( 1, K);P( K) >= 1 + ( N - 2) * X( K, 1));END用Lingo编的,条件是：从城市O出发经过城市A, B, C, D,E,F后又回到O的过程。

要求：编写程序,用回溯法求解旅行商问题。

} 四、实验结果
Backtrack(i+1); cw-=w[i];} if(cw+r>bestw){ x[i]=0; Backtrack(i+1);} r+=w[i]; } template<class Type> Type MaxLoading(Type w[],Type c,int n,int bestx[]) { Loading<Type>X; X.x=new int[n+1]; X.w=w; X.c=c; X.n=n; X.bestx=bestx; X.bestw=0; X.cw=0; X.r=0; for(int i=1;i<=n;i++) X.r+=w[i]; X.Backtrack(1); delete [] X.x; cout<<"所取物品："; for(i=1;i<=n;i++) cout<<bestx[i]<<" "; return X.bestw; } void main() { int w[100],c,n,beห้องสมุดไป่ตู้tx[6]; cout<<"输入物品个数(小于100)："; cin>>n; cout<<"输入"<<n<<"个物品重量："; for(int i=1;i<n+1;i++) cin>>w[i]; cout<<"输入第一艘轮船的载重量："; cin>>c; cout<<endl<<"最大装载重量为："<<MaxLoading(w,c,n,bestx)<<endl;

旅行商问题最简单的算法

+-------Program to solve travelling salesman problem------------------+example:[0]-------[1]| \4 /|| \ / |3| /\ |3| 6/ \ |[3]-------[2]1inputs:infinity:999no. of cities: 4no. of paths:6S D DistPath0:0 1 2Path1:0 2 4Path2:0 3 3Path3:1 2 3Path4:1 3 6Path5:2 3 1/////////////////////////////////////////////////////////////加了粗体和下滑线的是构环的问题#include<stdio.h>#include<conio.h>#define ALL -1#define MAXCITIES 10enum BOOL{FALSE,TRUE};long*visited; //visited nodes set herelong*min_circuit; //min inner circuit for given node as start node at position indexed 0 long*ham_circuit; //optimal circuit with length stored at position indexed 0long min_circuit_length; //min circuit lenth for given start nodeint n; //city countlong matrix[MAXCITIES][MAXCITIES]; //nondirectional nXn symmetric matrix///////////////to store path distances as sourceXdestinationlong INFI; // INFINITY value to be defined by user// function resets minimum circuit for a given start node///////////////////////////////////////// //with setting its id at index 0 and setting furthr node ids to -1void reset_min_circuit(int s_v_id){min_circuit[0]=s_v_id;for(int i=1;i<n;i++) min_circuit[i]=-1;}// marks given node id with given flag///////////////////////////////////////////////////////////// // if id==ALL it marks all nodes with given flagvoid set_visited(int v_id,BOOL flag){if(v_id==ALL) for(int i=0;i<n;i++) visited[i]=flag;else visited[v_id]=flag;}// function sets hamiltonion circuit for a given path length/////////////////////////////////////// //with setting it at index 0 and setting furthr nodes from current min_circuitvoid SET_HAM_CKT(long pl){ham_circuit[0]=pl;for(int i=0;i<n;i++) ham_circuit[i+1]=min_circuit[i];ham_circuit[n+1]=min_circuit[0];}//function sets a valid circuit by finiding min inner path for a given////////////////////////////// //combination start vertex and next vertex to start vertex such that// the 2nd vertex of circuits is always s_n_v and start and dest node is//always s_v for all possible values of s_n_v, and then returns the// valid circuit length for this combinationlong get_valid_circuit(int s_v,int s_n_v){int next_v,min,v_count=1;long path_length=0;min_circuit[0]=s_v;min_circuit[1]=s_n_v;set_visited(s_n_v,TRUE);path_length+=matrix[s_v][s_n_v];for(int V=s_n_v;v_count<n-1;v_count++){ min=INFI;for(int i=0;i<n;i++)if( matrix[V][i]<INFI && !visited[i]&& matrix[V][i]<=min)min=matrix[V][next_v=i];set_visited(next_v,TRUE);V=min_circuit[v_count+1]=next_v;path_length+=min;}path_length+=matrix[min_circuit[n-1]][s_v];return(path_length);}void main(){printf("Make sure that infinity value < sum of all path distances\nSet Infinity at (signed long):");scanf("%ld",&INFI);int pathcount,i,j,source,dest;long dist=0;long new_circuit_length=INFI;printf("Enter no. of cities(MAX:%d):",MAXCITIES);scanf("%d",&n);printf("Enter path count:");scanf("%d",&pathcount);printf("Enter paths:< source_id destination_id distance >\n ids varying from 0 to %d\n",n-1);//init all matrix distances to infinity//////for(i=0;i<n;i++)for(j=0;j<n;j++)matrix[i][j]=INFI;//populate the matrix///////////////for(i=0;i<pathcount;i++){printf("[path %d]:",i);scanf("%d %d %ld",&source,&dest,&dist);if(source!=dest)matrix[source][dest]=matrix[dest][source]=dist;}visited=new long[n];min_circuit=new long[n];ham_circuit=new long[n+2];min_circuit_length=INFI;int S_V_id=0;// algorithm//for each vertex, S_V as a staring nodefor(int S_V_id=0;S_V_id<n;S_V_id++){ //for each and non start vertex as ifor(i=0;i<n;i++){ //set all to unvisitedset_visited(ALL,FALSE);// set staring vertex as visitedset_visited(S_V_id,TRUE);//reset/init minimum circuitreset_min_circuit(S_V_id);// obtain circuit for combination of S_V and inew_circuit_length=get_valid_circuit(S_V_id,i);// if newer length is less than the previously//calculated min then set it as min and set the//current circuit in hamiltonion circuitif(new_circuit_length<=min_circuit_length)SET_HAM_CKT(min_circuit_length=new_circuit_length);}}// if any circuit found///////////////////if(min_circuit_length<INFI){printf("\n\nMinimum circuit length is: %ld\nCircuit is:\n",min_circuit_length);for(i=1;i<n+1;i++) printf("<%ld> ",ham_circuit[i]);}加环时，n+1要换成n+2else printf("\n\nNo hamiltonian circuit !");delete []visited;delete []min_circuit;delete []ham_circuit;}。

MATLAB多旅行商问题源代码

MATLAB多旅⾏商问题源代码MATLAB多旅⾏商问题源代码functionvarargout = mtspf_ga(xy,dmat,salesmen,min_tour,pop_size,num_iter,show_prog,show_res) % MTSPF_GA Fixed Multiple Traveling Salesmen Problem (M-TSP) Genetic Algorithm (GA)% Finds a (near) optimal solution to a variation of the M-TSP by setting% up a GA to search for the shortest route (least distance needed for% each salesman to travel from the start location to individual cities% and back to the original starting place)%% Summary:% 1. Each salesman starts at the first point, and ends at the first% point, but travels to a unique set of cities in between% 2. Except for the first, each city is visited by exactly one salesman%% Note: The Fixed Start/End location is taken to be the first XY point%% Input:% XY (float) is an Nx2 matrix of city locations, where N is the number of cities% DMAT (float) is an NxN matrix of city-to-city distances or costs% SALESMEN (scalar integer) is the number of salesmen to visit the cities% MIN_TOUR (scalar integer) is the minimum tour length for any of the% salesmen, NOT including the start/end point% POP_SIZE (scalar integer) is the size of the population (should be divisible by 8)% NUM_ITER (scalar integer) is the number of desired iterations for the algorithm to run% SHOW_PROG (scalar logical) shows the GA progress if true% SHOW_RES (scalar logical) shows the GA results if true%% Output:% OPT_RTE (integer array) is the best route found by the algorithm% OPT_BRK (integer array) is the list of route break points (these specify the indices% into the route used to obtain the individual salesman routes)% MIN_DIST (scalar float) is the total distance traveled by the salesmen%% Route/Breakpoint Details:% If there are 10 cities and 3 salesmen, a possible route/break% combination might be: rte = [5 6 9 4 2 8 10 3 7], brks = [3 7]% Taken together, these represent the solution [1 5 6 9 1][1 4 2 8 1][1 10 3 7 1],% which designates the routes for the 3 salesmen as follows:% . Salesman 1 travels from city 1 to 5 to 6 to 9 and back to 1% . Salesman 2 travels from city 1 to 4 to 2 to 8 and back to 1% . Salesman 3 travels from city 1 to 10 to 3 to 7 and back to 1%% 2D Example:% n = 35;% xy = 10*rand(n,2);% salesmen = 5;% min_tour = 3;% pop_size = 80;% num_iter = 5e3;% a = meshgrid(1:n);% dmat = reshape(sqrt(sum((xy(a,:)-xy(a',:)).^2,2)),n,n);% [opt_rte,opt_brk,min_dist] = mtspf_ga(xy,dmat,salesmen,min_tour, ... % pop_size,num_iter,1,1); %% 3D Example:% n = 35;% xyz = 10*rand(n,3);% salesmen = 5;% min_tour = 3;% pop_size = 80;% num_iter = 5e3;% a = meshgrid(1:n);% dmat = reshape(sqrt(sum((xyz(a,:)-xyz(a',:)).^2,2)),n,n);% [opt_rte,opt_brk,min_dist] = mtspf_ga(xyz,dmat,salesmen,min_tour, ... % pop_size,num_iter,1,1); %% See also: mtsp_ga, mtspo_ga, mtspof_ga, mtspofs_ga, mtspv_ga, distmat%% Author: Joseph Kirk% Email: jdkirk630@/doc/641797704.html% Release: 1.3% Release Date: 6/2/09% Process Inputs and Initialize Defaultsnargs = 8;for k = nargin:nargs-1switch kcase 0xy = 10*rand(40,2);case 1N = size(xy,1);a = meshgrid(1:N);dmat = reshape(sqrt(sum((xy(a,:)-xy(a',:)).^2,2)),N,N);case 2salesmen = 5;case 3min_tour = 2;case 4pop_size = 80;case 5num_iter = 5e3;case 6show_prog = 1;case 7show_res = 1;otherwiseendend% Verify Inputs[N,dims] = size(xy);[nr,nc] = size(dmat);if N ~= nr || N ~= ncerror('Invalid XY or DMAT inputs!')endn = N - 1; % Separate Start/End City% Sanity Checkssalesmen = max(1,min(n,round(real(salesmen(1)))));min_tour = max(1,min(floor(n/salesmen),round(real(min_tour(1))))); pop_size = max(8,8*ceil(pop_size(1)/8));num_iter = max(1,round(real(num_iter(1))));show_prog = logical(show_prog(1));show_res = logical(show_res(1));% Initializations for Route Break Point Selectionnum_brks = salesmen-1;dof = n - min_tour*salesmen; % degrees of freedom addto = ones(1,dof+1);for k = 2:num_brksaddto = cumsum(addto);endcum_prob = cumsum(addto)/sum(addto);% Initialize the Populationspop_rte = zeros(pop_size,n); % population of routespop_brk = zeros(pop_size,num_brks); % population of breaksfor k = 1:pop_sizepop_rte(k,:) = randperm(n)+1;pop_brk(k,:) = randbreaks();end% Select the Colors for the Plotted Routesclr = [1 0 0; 0 0 1; 0.67 0 1; 0 1 0; 1 0.5 0];if salesmen > 5clr = hsv(salesmen);end% Run the GAglobal_min = Inf;total_dist = zeros(1,pop_size);dist_history = zeros(1,num_iter);tmp_pop_rte = zeros(8,n);tmp_pop_brk = zeros(8,num_brks);new_pop_rte = zeros(pop_size,n);new_pop_brk = zeros(pop_size,num_brks);ifshow_progpfig = figure('Name','MTSPF_GA | Current Best Solution','Numbertitle','off'); end foriter = 1:num_iter% Evaluate Members of the Populationfor p = 1:pop_sizep_rte = pop_rte(p,:);p_brk = pop_brk(p,:);rng = [[1 p_brk+1];[p_brk n]]';for s = 1:salesmend = d + dmat(1,p_rte(rng(s,1))); % Add Start Distancefor k = rng(s,1):rng(s,2)-1d = d + dmat(p_rte(k),p_rte(k+1));endd = d + dmat(p_rte(rng(s,2)),1); % Add End Distanceendtotal_dist(p) = d;end% Find the Best Route in the Population[min_dist,index] = min(total_dist);dist_history(iter) = min_dist;ifmin_distglobal_min = min_dist;opt_rte = pop_rte(index,:);opt_brk = pop_brk(index,:);rng = [[1 opt_brk+1];[opt_brk n]]';ifshow_prog% Plot the Best Routefigure(pfig);for s = 1:salesmenrte = [1 opt_rte(rng(s,1):rng(s,2)) 1];if dims == 3, plot3(xy(rte,1),xy(rte,2),xy(rte,3),'.-','Color',clr(s,:)); else plot(xy(rte,1),xy(rte,2),'.-','Color',clr(s,:)); endtitle(sprintf('Total Distance = %1.4f, Iteration = %d',min_dist,iter)); hold onendif dims == 3, plot3(xy(1,1),xy(1,2),xy(1,3),'ko');else plot(xy(1,1),xy(1,2),'ko'); endhold offend% Genetic Algorithm Operatorsrand_grouping = randperm(pop_size);for p = 8:8:pop_sizertes = pop_rte(rand_grouping(p-7:p),:);brks = pop_brk(rand_grouping(p-7:p),:); dists = total_dist(rand_grouping(p-7:p)); [ignore,idx] = min(dists);best_of_8_rte = rtes(idx,:);best_of_8_brk = brks(idx,:);rte_ins_pts = sort(ceil(n*rand(1,2)));I = rte_ins_pts(1);J = rte_ins_pts(2);for k = 1:8 % Generate New Solutionstmp_pop_rte(k,:) = best_of_8_rte;tmp_pop_brk(k,:) = best_of_8_brk;switch kcase 2 % Fliptmp_pop_rte(k,I:J) = fliplr(tmp_pop_rte(k,I:J)); case 3 % Swaptmp_pop_rte(k,[I J]) = tmp_pop_rte(k,[J I]); case 4 % Slidetmp_pop_rte(k,I:J) = tmp_pop_rte(k,[I+1:J I]); case 5 % Modify Breakstmp_pop_brk(k,:) = randbreaks();case 6 % Flip, Modify Breakstmp_pop_rte(k,I:J) = fliplr(tmp_pop_rte(k,I:J)); tmp_pop_brk(k,:) = randbreaks();case 7 % Swap, Modify Breakstmp_pop_rte(k,[I J]) = tmp_pop_rte(k,[J I]); tmp_pop_brk(k,:) = randbreaks();case 8 % Slide, Modify Breakstmp_pop_rte(k,I:J) = tmp_pop_rte(k,[I+1:J I]); tmp_pop_brk(k,:) = randbreaks(); otherwise % Do Nothingendnew_pop_rte(p-7:p,:) = tmp_pop_rte;new_pop_brk(p-7:p,:) = tmp_pop_brk;endpop_rte = new_pop_rte;pop_brk = new_pop_brk;endifshow_res% Plotsfigure('Name','MTSPF_GA | Results','Numbertitle','off'); subplot(2,2,1);if dims == 3, plot3(xy(:,1),xy(:,2),xy(:,3),'k.');else plot(xy(:,1),xy(:,2),'k.'); endtitle('City Locations');subplot(2,2,2);imagesc(dmat([1 opt_rte],[1 opt_rte]));title('Distance Matrix');subplot(2,2,3);rng = [[1 opt_brk+1];[opt_brk n]]';for s = 1:salesmenrte = [1 opt_rte(rng(s,1):rng(s,2)) 1];if dims == 3, plot3(xy(rte,1),xy(rte,2),xy(rte,3),'.-','Color',clr(s,:)); else plot(xy(rte,1),xy(rte,2),'.-','Color',clr(s,:)); endtitle(sprintf('Total Distance = %1.4f',min_dist));hold on;endif dims == 3, plot3(xy(1,1),xy(1,2),xy(1,3),'ko');else plot(xy(1,1),xy(1,2),'ko'); endsubplot(2,2,4);plot(dist_history,'b','LineWidth',2);title('Best Solution History');set(gca,'XLim',[0 num_iter+1],'YLim',[0 1.1*max([1 dist_history])]); end% Return Outputsifnargoutvarargout{1} = opt_rte;varargout{2} = opt_brk;varargout{3} = min_dist;end% Generate Random Set of Break Pointsfunction breaks = randbreaks()ifmin_tour == 1 % No Constraints on Breakstmp_brks = randperm(n-1);breaks = sort(tmp_brks(1:num_brks));else % Force Breaks to be at Least the Minimum Tour Length num_adjust = find(rand < cum_prob,1)-1; spaces = ceil(num_brks*rand(1,num_adjust));adjust = zeros(1,num_brks);forkk = 1:num_brksadjust(kk) = sum(spaces == kk);endbreaks = min_tour*(1:num_brks) + cumsum(adjust);endendend。

蚁群算法旅行商问题代码

蚁群算法旅行商问题代码蚁群算法（Ant Colony Optimization, ACO）是一种基于蚁群行为的优化算法，常用于解决组合优化问题，如旅行商问题（Travelling Salesman Problem, TSP）。

下面是一个简单的Python 实现，使用蚁群算法解决TSP问题：```pythonimport numpy as npclass AntColony:def __init__(self, distances, n_ants, n_best, n_iteration, decay, alpha=1, beta=2): """Args:distances (2D numpy.array): Square matrix of distances. Diagonal is assumed to be np.inf.n_ants (int): Number of ants running per iterationn_best (int): Number of best ants who deposit pheromonen_iteration (int): Number of iterationsdecay (float): Rate it which pheromone decays. The pheromone value is multiplied by decay, so 0.95 will lead to decay, 0.5 to much faster decay.alpha (int or float): exponenet on pheromone, higher alpha gives pheromone more weight. Default=1beta (int or float): exponent on distance, higher beta give distance more weight. Default=2"""self.distances = distancesself.pheromone = np.ones(self.distances.shape) / len(distances)self.all_inds = range(len(distances))self.all_paths = self.gen_all_paths()self.n_ants = n_antsself.n_best = n_bestself.n_iteration = n_iterationself.decay = decayself.alpha = alphaself.beta = betadef gen_all_paths(self):all_paths = []for i in self.all_inds:rest = set(self.all_inds)current = []rest.remove(i)for _ in range(len(self.distances)-1):to_visit = list(rest)probs = self.pheromone[i, to_visit]**self.alpha * ((1.0 / self.distances[i, to_visit])**self.beta)probs /= sum(probs)next_ind = np.random.choice(to_visit, p=probs)current.append((i, next_ind))i = next_indrest.remove(next_ind)all_paths.append(current)return all_pathsdef gen_path_dist(self, path):total_dist = 0for ant in path:total_dist += self.distances[ant]return total_distdef run(self):all_time_best_path = Noneall_time_best_dist = np.inffor i in range(self.n_iteration):all_paths = self.gen_all_paths()self.spread_pheronome(all_paths, self.n_best, self.distances)self.pheromone * self.decayif self.gen_path_dist(all_paths[0]) < all_time_best_dist:all_time_best_path = all_paths[0]all_time_best_dist = self.gen_path_dist(all_paths[0])self.global_best_path_ = all_time_best_pathself.global_best_dist_ = all_time_best_distreturn all_time_best_pathdef spread_pheronome(self, all_paths, n_best, dists):sorted_paths = sorted(all_paths, key=lambda x: self.gen_path_dist(x))for path in sorted_paths[:n_best]:for move in path:self.pheromone[move] += 1.0 / dists[move]# Example Usage:# Define distances between cities (replace this with your own data)distances = np.array([[np.inf, 2, 2, 5, 7],[2, np.inf, 4, 8, 2],[2, 4, np.inf, 1, 3],[5, 8, 1, np.inf, 2],[7, 2, 3, 2, np.inf]])# Create an AntColony instanceant_colony = AntColony(distances, n_ants=5, n_best=2, n_iteration=100, decay=0.95, alpha=1, beta=2)# Run the algorithmbest_path = ant_colony.run()print("Best Path:", best_path)print("Best Distance:", ant_colony.global_best_dist_)```这个示例中，`distances` 表示城市之间的距离矩阵。

旅行售货员问题c++源代码及运行结果

//源代码：#include <stdio.h>#include <malloc.h>#define NoEdge 1000struct MinHeapNode{int lcost; //子树费用的下界int cc; //当前费用int rcost; //x[s:n-1]中顶点最小出边费用和int s; //根节点到当前节点的路径为x[0:s]int *x; //需要进一步搜索的顶点是//x[s+1:n-1]struct MinHeapNode *next;};int n; //图G的顶点数int **a; //图G的邻接矩阵//int NoEdge; //图G的无边标记int cc; //当前费用int bestc; //当前最小费用MinHeapNode* head = 0; /*堆头*/MinHeapNode* lq = 0; /*堆第一个元素*/ MinHeapNode* fq = 0; /*堆最后一个元素*/int DeleteMin(MinHeapNode*&E){MinHeapNode* tmp = NULL;tmp = fq;// w = fq->weight ;E = fq;if(E == NULL)return 0;head->next = fq->next; /*一定不能丢了链表头*/fq = fq->next;// free(tmp) ;return 0;}int Insert(MinHeapNode* hn){if(head->next == NULL){head->next = hn; //将元素放入链表中fq = lq = head->next; //一定要使元素放到链中}else{MinHeapNode *tmp = NULL;tmp = fq;if(tmp->cc > hn->cc){hn->next = tmp;head->next = hn;fq = head->next; /*链表只有一个元素的情况*/ }else{for(; tmp != NULL;){if(tmp->next != NULL && tmp->cc > hn->cc){hn->next = tmp->next;tmp->next = hn;break;}tmp = tmp->next;}}if(tmp == NULL){lq->next = hn;lq = lq->next;}}return 0;}int BBTSP(int v[]){//解旅行售货员问题的优先队列式分支限界法/*初始化最优队列的头结点*/head = (MinHeapNode*)malloc(sizeof(MinHeapNode));head->cc = 0;head->x = 0;head->lcost = 0;head->next = NULL;head->rcost = 0;head->s = 0;int *MinOut = new int[n + 1]; /*定义定点i的最小出边费用*/ //计算MinOut[i]=顶点i的最小出边费用int MinSum = 0;//最小出边费用总合for(int i = 1; i <= n; i++){int Min = NoEdge; /*定义当前最小值*/for(int j = 1; j <= n; j++)if(a[i][j] != NoEdge && /*当定点i,j之间存在回路时*/(a[i][j] < Min || Min == NoEdge)) /*当顶点i,j之间的距离小于Min*/Min = a[i][j]; /*更新当前最小值*/if(Min == NoEdge)return NoEdge;//无回路MinOut[i] = Min; /*顶点i的最小出边费用*/MinSum += Min; /*最小出边费用的总和*/}MinHeapNode *E = 0;E = (MinHeapNode*)malloc(sizeof(MinHeapNode));E->x = new int[n];// E.x=new int[n];for(i = 0; i < n; i++)E->x[i] = i + 1;E->s = 0;E->cc = 0;E->rcost = MinSum;E->next = 0; //初始化当前扩展节点int bestc = NoEdge; /*记录当前最小值*///搜索排列空间树while(E->s < n - 1){//非叶结点if(E->s == n - 2){//当前扩展结点是叶结点的父结点if(a[E->x[n - 2]][E->x[n - 1]] != NoEdge && /*当前要扩展和叶节点有边存在*/ a[E->x[n - 1]][1] != NoEdge && /*当前页节点有回路*/(E->cc + a[E->x[n - 2]][E->x[n - 1]] + a[E->x[n - 1]][1] < bestc /*该节点相应费用小于最小费用*/|| bestc == NoEdge)){bestc = E->cc + a[E->x[n - 2]][E->x[n - 1]] + a[E->x[n - 1]][1]; /*更新当前最新费用*/E->cc = bestc;E->lcost = bestc;E->s++;E->next = NULL;Insert(E); /*将该页节点插入到优先队列中*/}elsefree(E->x);//该页节点不满足条件舍弃扩展结点}else{//产生当前扩展结点的儿子结点for(i = E->s + 1; i < n; i++)if(a[E->x[E->s]][E->x[i]] != NoEdge){ /*当前扩展节点到其他节点有边存在*///可行儿子结点int cc = E->cc + a[E->x[E->s]][E->x[i]]; /*加上节点i后当前节点路径*/int rcost = E->rcost - MinOut[E->x[E->s]]; /*剩余节点的和*/int b = cc + rcost; //下界if(b < bestc || bestc == NoEdge){//子树可能含最优解,结点插入最小堆MinHeapNode * N;N = (MinHeapNode*)malloc(sizeof(MinHeapNode));N->x = new int[n];for(int j = 0; j < n; j++)N->x[j] = E->x[j];N->x[E->s + 1] = E->x[i];N->x[i] = E->x[E->s + 1];/*添加当前路径*/N->cc = cc; /*更新当前路径距离*/N->s = E->s + 1; /*更新当前节点*/N->lcost = b; /*更新当前下界*/N->rcost = rcost;N->next = NULL;Insert(N); /*将这个可行儿子结点插入到活结点优先队列中*/}}free(E->x);}//完成结点扩展DeleteMin(E);//取下一扩展结点if(E == NULL)break; //堆已空}if(bestc == NoEdge)return NoEdge;//无回路for(i = 0; i < n; i++)v[i + 1] = E->x[i];//将最优解复制到v[1:n]while(true){//释放最小堆中所有结点free(E->x);DeleteMin(E);if(E == NULL)break;}return bestc;}int main(){n = 0;int i = 0;scanf("%d", &n);a = (int**)malloc(sizeof(int*) * (n + 1));for(i = 1; i <= n; i++){a[i] = (int*)malloc(sizeof(int) * (n + 1));}for(i = 1; i <= n; i++)for(int j = 1; j <= n; j++)scanf("%d", &a[i][j]);// prev = (int*)malloc(sizeof(int)*(n+1)) ;int*v = (int*)malloc(sizeof(int) * (n + 1));// MaxLoading(w , c , n) ;for(i = 1; i <= n; i++)v[i] = 0;bestc = BBTSP(v);for(i = 1; i <= n; i++)printf("%d\t", v[i]);printf("\n");printf("%d\n", bestc);return 0;}运行结果：。

动态规划解决旅行商问题_附代码

1.问题基本描述:求一个旅行商经过N个城市最后回到出发点的最短路径.即,在一个无向带权图的邻接矩阵中,求一个最短环包括所有顶点.2.解法:1)动态规划:假设从顶点i出发，令d(i,V’)表示从顶点i出发经过V’中各个顶点一次且仅一次，最后回到出发点i的最短路径的长度，开始时，V’=V-{i}，于是，旅行商问题的动态规划函数为：d(i,V’) = min{c ik + d(k,V’-{k})} (k∈V’)1)d(k,{}) = c ki (k ≠ i)2)简单来说,就是用递归表达:从出发点0到1号点,假设1是第一个,则剩下的路程就是从1经过剩下的点最后回到0点的最短路径. 所以当V’为空的时候, d(k,{}) = c ki (k ≠ i), 找的是最后一个点到0点的距离.递归求解1之后,再继续求V’之中剩下的点,最后找出min.如果按照这个思想直接做,对于每一个i都要递归剩下的V中所有的点,所以这样的时间复杂度就近似于N!,其中有很多重复的工作.可以从小的集合到大的集合算,并存入一个二维数组,这样当加入一个节点时,就可以用到之前的结果,如四个点的情况:动态填表:表中元素代表第i个节点经过V集合中的点最后到0点的最短值.如果有多个值,取其中最小的一个.这样一共循环(2^(N-1)-1)*(N-1)次,就把时间复杂度缩小到O(N*2)的级别.核心伪代码如下:{for (i =1;i<n;i++) //初始化第0列d[i][0]=c[i][0];for( j=1;j<2^(N-1)-1;j++)for(i=1 ; i<n ;i++){if(子集Vj中不包含i){对Vj中的每个元素k，计算d[i][Vj] = min{c[i][k] + d[k][{Vj-k}] | 每一个k∈Vj};}}对V[2^(n-1)-1]中的每个元素k,计算:d[0][2^(n-1)-1] = min{c[0][k] + d[k][2^(n-1)-2]};输出最短路径:d[0][2^(n-1)-1];}。

旅行商问题JAVA代码

import java.util.*;public class Tsp {private String cityName[]={"北京","上海","天津","重庆","哈尔滨","长春","沈阳","呼和浩特","石家庄","太原","济南","郑州","西安","兰州","银川","西宁","乌鲁木齐","合肥","南京","杭州","长沙","南昌","武汉","成都","贵州","福建","台北","广州","海口","南宁","昆明","拉萨","香港","澳门"};//private String cityEnd[]=new String[34];private int cityNum=cityName.length; //城市个数private int popSize = 50; //种群数量private int maxgens = 20000; //迭代次数private double pxover = 0.8; //交叉概率private double pmultation = 0.05; //变异概率private long[][] distance = new long[cityNum][cityNum];private int range = 2000; //用于判断何时停止的数组区间private class genotype {int city[] = new int[cityNum]; //单个基因的城市序列long fitness; //该基因的适应度double selectP; //选择概率double exceptp; //期望概率int isSelected; //是否被选择}private genotype[] citys = new genotype[popSize];/*** 构造函数，初始化种群*/public Tsp() {for (int i = 0; i < popSize; i++) {citys[i] = new genotype();int[] num = new int[cityNum];for (int j = 0; j < cityNum; j++)num[j] = j;int temp = cityNum;for (int j = 0; j < cityNum; j++) {int r = (int) (Math.random() * temp);citys[i].city[j] = num[r];num[r] = num[temp - 1];temp--;}citys[i].fitness = 0;citys[i].selectP = 0;citys[i].exceptp = 0;citys[i].isSelected = 0;}initDistance();}/*** 计算每个种群每个基因个体的适应度，选择概率，期望概率，和是否被选择。

旅行商问题编码解读

旅行商问题编码解读
旅行商问题是一个经典的组合优化问题，其目标是找到一条路线，使得旅行商可以访问给定的一组城市，并回到起点城市，同
时让旅行路径的总长度最小。

该问题在现实生活中有许多应用，
如物流规划、电路设计等。

为了解决这个问题，可以使用编码来表示旅行路径。

一种常见
的编码方式是采用序列编码，即将路径表示为一串数字，每个数
字代表一个城市的访问顺序。

例如，一个表示四个城市的路径可
以被编码为"1-3-2-4-1"，其中数字表示城市的标识符。

在编码过程中，需要考虑以下几个要点。

首先，路径的起点和
终点是相同的，旅行商必须回到起点城市。

其次，每个城市只能
访问一次，确保路径的完整性和合法性。

最后，路径的总长度应
该最小化。

为了找到最佳路径，可以使用不同的算法和技术。

其中，著名
的算法包括贪婪算法、遗传算法和蚁群算法等。

这些算法通过迭
代和优化的过程，逐步寻找最优解。

同时，还可以使用启发式方
法来加速搜索过程，如局部搜索和模拟退火等。

旅行商问题的解决有助于提高资源利用和路径规划的效率，从
而优化物流配送、线路规划等方面的问题。

通过合理地编码和应
用有效的算法，可以更好地解决旅行商问题，并应用于实际场景中。

旅行商问题是一个重要而具有挑战性的问题，通过编码和算法
的应用，可以找到最优的旅行路径。

这有助于解决各种实际问题，并提高资源利用和路线规划的效率。

蚁群算法旅行商问题人工智能 java 代码+报告+注释

目录1. 整体结构与功能划分 (2)(一) 总体结构 (2)(二) 类的划分与功能说明 (2)i. NewJFrame 类 (2)ii. Ant类 (2)iii. ACO类 (2)iv. parameter类 (3)v. Data类 (3)2. 输入输出格式 (3)(一) 输入信息 (3)i. 城市坐标信息 (3)ii. 参数 (3)(二) 输出信息 (3)i. 图形用户界面输出 (3)ii. 控制台输出 (3)3. 完整代码(见附录) (4)4. 算法优化 (4)（一）蚁恒模型 (4)（二）信息素强度及路径信息素初始值 (4)（三）各参数对算法的影响 (4)5. 运行测试 (5)(一) 传统模型 (5)(二) 蚁恒模型 (6)6. 参考文献 (7)附录 (7)(一) NewJFrame.java (7)(二) Ant.java (13)(三) ACO.java (17)(四) Parameter.java (23)(五) Data.java (23)(六) NewJFramePath.java (24)1.整体结构与功能划分(一)总体结构程序的主类是NewJFrame，该类负责绘制图形用户界面并执行蚁群算法。

将蚁群算法中的“蚂蚁”抽象为Ant类，实现遍历所有城市（选择下一个城市）、计算本次遍历的路程等功能。

ACO类聚合了Ant类，完整的实现了蚁群算法，包括迭代遍历、更新信息素等过程。

NewJFramePath类将解路径以图形方式显示出来。

另设两个数据类parameter和Data，parameter类用于保存参数的默认值，主要参数包括：蚂蚁数量、城市数目、迭代次数、α、β、挥发因子、信息素强度Q、路径信息素初始值initQ。

Data中存储每个城市的位置信息。

(二)类的划分与功能说明i.NewJFrame类定义了各组件对鼠标事件响应方式。

图形用户界面为用户提供输入各项参数的窗口，并添加了“使用默认参数”的功能。

旅行商问题C程序

旅行商问题C程序function varargout = tsp_ga(xy,dmat,pop_size,num_iter,show_prog,show_res)nargs = 6;for k = nargin:nargs-1switch kcase 0xy = 10*rand(50,2);case 1N = size(xy,1);a = meshgrid(1:N);dmat = reshape(sqrt(sum((xy(a,:)-xy(a',:)).^2,2)),N,N);case 2pop_size = 100;case 3num_iter = 1e4;case 4show_prog = 1;case 5show_res = 1;otherwiseendend% Verify Inputs[N,dims] = size(xy);[nr,nc] = size(dmat);if N ~= nr || N ~= ncerror('Invalid XY or DMAT inputs!') endn = N;% Sanity Checkspop_size = 4*ceil(pop_size/4);num_iter = max(1,round(real(num_iter(1)))); show_prog = logical(show_prog(1));show_res = logical(show_res(1));% Initialize the Populationpop = zeros(pop_size,n);for k = 1:pop_sizepop(k,:) = randperm(n);end% Run the GAglobal_min = Inf;total_dist = zeros(1,pop_size);dist_history = zeros(1,num_iter);tmp_pop = zeros(4,n);new_pop = zeros(pop_size,n);if show_progpfig = figure('Name','TSP_GA | Current Best Solution','Numbertitle','off');endfor iter = 1:num_iter% Evaluate Each Population Member (Calculate Total Distance)for p = 1:pop_sized = dmat(pop(p,n),pop(p,1)); % Closed Pathfor k = 2:nd = d + dmat(pop(p,k-1),pop(p,k));endtotal_dist(p) = d;end% Find the Best Route in the Population[min_dist,index] = min(total_dist);dist_history(iter) = min_dist;if min_dist < global_minglobal_min = min_dist;opt_rte = pop(index,:);if show_prog% Plot the Best Routefigure(pfig);rte = opt_rte([1:n 1]);if dims == 3, plot3(xy(rte,1),xy(rte,2),xy(rte,3),'r.-');else plot(xy(rte,1),xy(rte,2),'r.-'); endtitle(sprintf('Total Distance = %1.4f, Iteration = %d',min_dist,iter)); endend% Genetic Algorithm Operatorsrand_pair = randperm(pop_size);for p = 4:4:pop_sizertes = pop(rand_pair(p-3:p),:);dists = total_dist(rand_pair(p-3:p));[ignore,idx] = min(dists);best_of_4_rte = rtes(idx,:);ins_pts = sort(ceil(n*rand(1,2)));I = ins_pts(1);J = ins_pts(2);for k = 1:4 % Mutate the Best to get Three New Routes tmp_pop(k,:) = best_of_4_rte;switch kcase 2 % Fliptmp_pop(k,I:J) = fliplr(tmp_pop(k,I:J));case 3 % Swaptmp_pop(k,[I J]) = tmp_pop(k,[J I]);case 4 % Slidetmp_pop(k,I:J) = tmp_pop(k,[I+1:J I]);otherwise % Do Nothingendendnew_pop(p-3:p,:) = tmp_pop;endpop = new_pop;endif show_res% Plots the GA Resultsfigure('Name','TSP_GA | Results','Numbertitle','off');subplot(2,2,1);if dims == 3, plot3(xy(:,1),xy(:,2),xy(:,3),'k.');else plot(xy(:,1),xy(:,2),'k.'); endtitle('City Locations');subplot(2,2,2);imagesc(dmat(opt_rte,opt_rte));title('Distance Matrix');subplot(2,2,3);rte = opt_rte([1:n 1]);if dims == 3, plot3(xy(rte,1),xy(rte,2),xy(rte,3),'r.-');else plot(xy(rte,1),xy(rte,2),'r.-'); endtitle(sprintf('Total Distance = %1.4f',min_dist));subplot(2,2,4);plot(dist_history,'b','LineWidth',2);title('Best Solution History');set(gca,'XLim',[0 num_iter+1],'YLim',[0 1.1*max([1 dist_history])]);end% Return Outputsif nargoutvarargout{1} = opt_rte; varargout{2} = min_dist; end。

旅行商问题

#include<stdio.h>#include<conio.h>int c = -1, dist[20][20], n, artemp[20];typedef struct {int a[20];int sum;int pro;int k;int parent;int cost;int min;} G;G set[100], temp, search;int find(int pro, int sum, int p){int i;for (i = 0; i <= c; i++)if (set[i].pro == pro && set[i].sum == sum && set[i].parent == p)return i;return 0;}void cost(int pro, int sum, int p, int k){int i, ar[20], min, loc, temploc, el;temploc = find(pro, sum, p);el = set[temploc].a[0];if (k == 1){set[temploc].cost = dist[p - 1][el - 1] + dist[el - 1][0];set[temploc].min = el;}else{loc = find(pro / el, sum - el, el);set[temploc].cost = ar[0] = min = set[find(pro / el, sum - el, el)].cost + dist[p - 1][el - 1];set[temploc].min = el;for (i = 1; i < k; i++){loc = find(pro / set[temploc].a[i], sum - set[temploc].a[i], set[temploc].a[i]);ar[i] = set[loc].cost + dist[p-1][set[temploc].a[i] - 1];if (ar[i] < min){set[temploc].cost = min = ar[i];set[temploc].min = set[temploc].a[i];}}}}void init(int k, int parent){temp.sum = 0;temp.pro = 1;temp.k = k;temp.parent = temp.min = parent;}void sort(){int i, j;for (i = 0; i < c; i++)for (j = 0; j < c; j++)if (set[j + 1].sum < set[j].sum){temp = set[j];set[j] = set[j + 1];set[j + 1] = temp;}}void fill(int parent, int ar[], int k){int i, j, counter, tempar[20], p;for (i = 0; i <= k; i++){init(k, 1);counter = -1;for (j = 0; j <= k; j++)if (i != j){temp.a[++counter] = ar[j];tempar[counter] = ar[j];temp.pro *= ar[j];temp.sum += ar[j];}elsep = ar[j];if (!find(temp.pro, temp.sum, p)){temp.parent = temp.min = p;if (k>=0)set[++c] = temp;fill(p, tempar, k - 1);}if (!find(temp.pro, temp.sum, parent) && parent != 1){temp.parent = temp.min = parent;if (k>=0)set[++c] = temp;fill(parent, tempar, k - 1);}}}void genpath(int t[]){int i = 0, ctemp=c;t[i++] = set[ctemp].min;while (i < n){ctemp = find(set[ctemp].pro / t[i - 1], set[ctemp].sum - t[i - 1], t[i - 1]);t[i++] = set[ctemp].min;}}void main(){int i, j, l, pro, sum, path[20];printf("Enter number of cities: ");scanf("%d", &n);init(n - 1, 1);for (i = 2; i <= n; i++){temp.a[i - 2] = i;artemp[i - 2] = i;temp.sum += i;temp.pro *= i;}pro = temp.pro;sum = temp.sum;set[++c] = temp;fill(1, artemp, n - 2);sort();printf("\n");for (i = 0; i < n; i++)for (j = 0; j < n; j++){scanf("%d", &dist[i][j]);if (j == 0 && i != 0)set[find(1, 0, i+1)].cost = dist[i][0];}for (i = n-1; i <= c; i++)cost(set[i].pro, set[i].sum, set[i].parent, set[i].k); printf("\n\n");for (i = 0; i <= c; i++){printf("g(%d, ", set[i].parent);if (set[i].k > 0){printf("{%d", set[i].a[0]);for (j = 1; j < set[i].k; j++)printf(", %d", set[i].a[j]);printf("}");}printf(") = %d ", set[i].cost);for (l = 0; l < set[c].k - set[i].k/2; l++)printf(" ");printf("p(%d, ", set[i].parent);if (set[i].k > 0){printf("{%d", set[i].a[0]);for (j = 1; j < set[i].k; j++)printf(", %d ", set[i].a[j]);printf("}");}printf(") = %d\n\n", set[i].min);}genpath(path);printf("\nDistance travelled: %d\n", set[c].cost); printf("\n\nPath: 1 --> ");for (i = 0; i < n - 1; i++)printf("%d --> ", path[i]);printf("1");getch();}/*Enter number of cities: 40 2 9 101 0 6 415 7 0 86 3 12 0g(4, ) = 6 p(4, ) = 4g(3, ) = 15 p(3, ) = 3g(2, ) = 1 p(2, ) = 2g(4, {2}) = 4 p(4, {2}) = 2g(3, {2}) = 8 p(3, {2}) = 2g(4, {3}) = 27 p(4, {3}) = 3g(2, {3}) = 21 p(2, {3}) = 3g(3, {4}) = 14 p(3, {4}) = 4g(2, {4}) = 10 p(2, {4}) = 4g(4, {2, 3}) = 20 p(4, {2, 3}) = 3 g(3, {2, 4}) = 12 p(3, {2, 4}) = 4 g(2, {3, 4}) = 20 p(2, {3, 4}) = 3 g(1, {2, 3, 4}) = 21 p(1, {2, 3 , 4}) = 3 Distance travelled: 21Path: 1 --> 3 --> 4 --> 2 --> 1 */。

LINGO软件求旅行商(货郎担)问题的程序与答案

PABCD P-A-C-B-D-Pdist=0 70 115 90 9570 0 46 21 50115 46 0 30 3290 21 30 0 4895 50 32 48 0;Global optimal solution found.Objective value: 248.0000Extended solver steps: 0Total solver iterations: 36Variable Value Reduced Cost N 5.000000 0.000000 U( 1) 0.000000 0.000000 U( 2) 3.000000 0.000000 U( 3) 1.000000 0.000000U( 5) 0.000000 0.000000 DIST( 1, 1) 0.000000 0.000000 DIST( 1, 2) 70.00000 0.000000 DIST( 1, 3) 115.0000 0.000000 DIST( 1, 4) 90.00000 0.000000 DIST( 1, 5) 95.00000 0.000000 DIST( 2, 1) 70.00000 0.000000 DIST( 2, 2) 0.000000 0.000000 DIST( 2, 3) 46.00000 0.000000 DIST( 2, 4) 21.00000 0.000000 DIST( 2, 5) 50.00000 0.000000 DIST( 3, 1) 115.0000 0.000000 DIST( 3, 2) 46.00000 0.000000 DIST( 3, 3) 0.000000 0.000000 DIST( 3, 4) 30.00000 0.000000 DIST( 3, 5) 32.00000 0.000000 DIST( 4, 1) 90.00000 0.000000 DIST( 4, 2) 21.00000 0.000000 DIST( 4, 3) 30.00000 0.000000 DIST( 4, 4) 0.000000 0.000000 DIST( 4, 5) 48.00000 0.000000 DIST( 5, 1) 95.00000 0.000000 DIST( 5, 2) 50.00000 0.000000 DIST( 5, 3) 32.00000 0.000000 DIST( 5, 4) 48.00000 0.000000 DIST( 5, 5) 0.000000 0.000000 X( 1, 1) 0.000000 0.000000 X( 1, 2) 0.000000 70.00000 X( 1, 3) 0.000000 115.0000 X( 1, 4) 0.000000 90.00000 X( 1, 5) 1.000000 95.00000 X( 2, 1) 1.000000 70.00000 X( 2, 2) 0.000000 0.000000 X( 2, 3) 0.000000 46.00000 X( 2, 4) 0.000000 21.00000 X( 2, 5) 0.000000 50.00000 X( 3, 1) 0.000000 115.0000 X( 3, 2) 0.000000 46.00000 X( 3, 3) 0.000000 0.000000 X( 3, 4) 1.000000 30.00000 X( 3, 5) 0.000000 32.00000 X( 4, 1) 0.000000 90.00000 X( 4, 2) 1.000000 21.00000X( 4, 4) 0.000000 0.000000 X( 4, 5) 0.000000 48.00000 X( 5, 1) 0.000000 95.00000 X( 5, 2) 0.000000 50.00000 X( 5, 3) 1.000000 32.00000 X( 5, 4) 0.000000 48.00000 X( 5, 5) 0.000000 0.000000Row Slack or Surplus Dual Price1 0.000000 0.0000002 248.0000 -1.0000003 0.000000 0.0000004 0.000000 0.0000005 0.000000 0.0000006 0.000000 0.0000007 0.000000 0.0000008 0.000000 0.0000009 0.000000 0.00000010 0.000000 0.00000011 0.000000 0.00000012 0.000000 0.00000013 2.000000 0.00000014 3.000000 0.00000015 1.000000 0.00000016 6.000000 0.00000017 0.000000 0.00000018 3.000000 0.00000019 0.000000 0.00000020 3.000000 0.00000021 2.000000 0.00000022 7.000000 0.00000023 0.000000 0.00000024 6.000000 0.00000025 0.000000 0.00000026 2.000000 0.00000027 1.000000 0.00000028 3.000000 0.000000 PABCE P-E-B-C-A-Pdist=0 70 115 90 8570 0 46 21 60115 46 0 30 5590 21 30 0 5385 60 55 53 0;Global optimal solution found.Objective value: 261.0000Extended solver steps: 0Total solver iterations: 10Variable Value Reduced Cost N 5.000000 0.000000 U( 1) 0.000000 0.000000 U( 2) 3.000000 0.000000 U( 3) 1.000000 0.000000 U( 4) 2.000000 0.000000DIST( 1, 1) 0.000000 0.000000 DIST( 1, 2) 70.00000 0.000000 DIST( 1, 3) 115.0000 0.000000 DIST( 1, 4) 90.00000 0.000000 DIST( 1, 5) 85.00000 0.000000 DIST( 2, 1) 70.00000 0.000000 DIST( 2, 2) 0.000000 0.000000 DIST( 2, 3) 46.00000 0.000000 DIST( 2, 4) 21.00000 0.000000 DIST( 2, 5) 60.00000 0.000000 DIST( 3, 1) 115.0000 0.000000 DIST( 3, 2) 46.00000 0.000000 DIST( 3, 3) 0.000000 0.000000 DIST( 3, 4) 30.00000 0.000000 DIST( 3, 5) 55.00000 0.000000 DIST( 4, 1) 90.00000 0.000000 DIST( 4, 2) 21.00000 0.000000 DIST( 4, 3) 30.00000 0.000000 DIST( 4, 4) 0.000000 0.000000 DIST( 4, 5) 53.00000 0.000000 DIST( 5, 1) 85.00000 0.000000 DIST( 5, 2) 60.00000 0.000000 DIST( 5, 3) 55.00000 0.000000 DIST( 5, 4) 53.00000 0.000000 DIST( 5, 5) 0.000000 0.000000 X( 1, 1) 0.000000 0.000000 X( 1, 2) 0.000000 70.00000 X( 1, 3) 0.000000 115.0000 X( 1, 4) 0.000000 90.00000 X( 1, 5) 1.000000 85.00000 X( 2, 1) 1.000000 70.00000 X( 2, 2) 0.000000 0.000000 X( 2, 3) 0.000000 46.00000 X( 2, 4) 0.000000 21.00000 X( 2, 5) 0.000000 60.00000 X( 3, 1) 0.000000 115.0000 X( 3, 2) 0.000000 46.00000 X( 3, 3) 0.000000 0.000000 X( 3, 4) 1.000000 30.00000 X( 3, 5) 0.000000 55.00000 X( 4, 1) 0.000000 90.00000 X( 4, 2) 1.000000 21.00000 X( 4, 3) 0.000000 30.00000X( 4, 5) 0.000000 53.00000 X( 5, 1) 0.000000 85.00000 X( 5, 2) 0.000000 60.00000 X( 5, 3) 1.000000 55.00000 X( 5, 4) 0.000000 53.00000 X( 5, 5) 0.000000 0.000000Row Slack or Surplus Dual Price1 0.000000 0.0000002 261.0000 -1.0000003 0.000000 0.0000004 0.000000 0.0000005 0.000000 0.0000006 0.000000 0.0000007 0.000000 0.0000008 0.000000 0.0000009 0.000000 0.00000010 0.000000 0.00000011 0.000000 0.00000012 0.000000 0.00000013 2.000000 0.00000014 3.000000 0.00000015 1.000000 0.00000016 6.000000 0.00000017 0.000000 0.00000018 3.000000 0.00000019 0.000000 0.00000020 3.000000 0.00000021 2.000000 0.00000022 7.000000 0.00000023 0.000000 0.00000024 6.000000 0.00000025 0.000000 0.00000026 2.000000 0.00000027 1.000000 0.00000028 3.000000 0.000000 PABDE P-E-D-B-A-Pdist=0 70 115 95 8570 0 46 50 60115 46 0 32 5595 50 32 0 2185 60 55 21 0;Global optimal solution found.Objective value: 254.0000Extended solver steps: 0Total solver iterations: 24Variable Value Reduced Cost N 5.000000 0.000000 U( 1) 0.000000 0.000000 U( 2) 3.000000 0.000000 U( 3) 2.000000 0.000000 U( 4) 1.000000 0.000000 U( 5) 0.000000 0.000000 DIST( 1, 1) 0.000000 0.000000DIST( 1, 3) 115.0000 0.000000 DIST( 1, 4) 95.00000 0.000000 DIST( 1, 5) 85.00000 0.000000 DIST( 2, 1) 70.00000 0.000000 DIST( 2, 2) 0.000000 0.000000 DIST( 2, 3) 46.00000 0.000000 DIST( 2, 4) 50.00000 0.000000 DIST( 2, 5) 60.00000 0.000000 DIST( 3, 1) 115.0000 0.000000 DIST( 3, 2) 46.00000 0.000000 DIST( 3, 3) 0.000000 0.000000 DIST( 3, 4) 32.00000 0.000000 DIST( 3, 5) 55.00000 0.000000 DIST( 4, 1) 95.00000 0.000000 DIST( 4, 2) 50.00000 0.000000 DIST( 4, 3) 32.00000 0.000000 DIST( 4, 4) 0.000000 0.000000 DIST( 4, 5) 21.00000 0.000000 DIST( 5, 1) 85.00000 0.000000 DIST( 5, 2) 60.00000 0.000000 DIST( 5, 3) 55.00000 0.000000 DIST( 5, 4) 21.00000 0.000000 DIST( 5, 5) 0.000000 0.000000 X( 1, 1) 0.000000 0.000000 X( 1, 2) 0.000000 70.00000 X( 1, 3) 0.000000 115.0000 X( 1, 4) 0.000000 95.00000 X( 1, 5) 1.000000 85.00000 X( 2, 1) 1.000000 70.00000 X( 2, 2) 0.000000 0.000000 X( 2, 3) 0.000000 46.00000 X( 2, 4) 0.000000 50.00000 X( 2, 5) 0.000000 60.00000 X( 3, 1) 0.000000 115.0000 X( 3, 2) 1.000000 46.00000 X( 3, 3) 0.000000 0.000000 X( 3, 4) 0.000000 32.00000 X( 3, 5) 0.000000 55.00000 X( 4, 1) 0.000000 95.00000 X( 4, 2) 0.000000 50.00000 X( 4, 3) 1.000000 32.00000 X( 4, 4) 0.000000 0.000000 X( 4, 5) 0.000000 21.00000X( 5, 2) 0.000000 60.00000 X( 5, 3) 0.000000 55.00000 X( 5, 4) 1.000000 21.00000 X( 5, 5) 0.000000 0.000000Row Slack or Surplus Dual Price1 0.000000 0.0000002 254.0000 -1.0000003 0.000000 0.0000004 0.000000 0.0000005 0.000000 0.0000006 0.000000 0.0000007 0.000000 0.0000008 0.000000 0.0000009 0.000000 0.00000010 0.000000 0.00000011 0.000000 0.00000012 0.000000 0.00000013 3.000000 0.00000014 2.000000 0.00000015 1.000000 0.00000016 0.000000 0.00000017 3.000000 0.00000018 2.000000 0.00000019 6.000000 0.00000020 0.000000 0.00000021 3.000000 0.00000022 7.000000 0.00000023 6.000000 0.00000024 0.000000 0.00000025 0.000000 0.00000026 1.000000 0.00000027 2.000000 0.00000028 3.000000 0.000000 PACDE P-A-C-D-E-Pdist=0 70 90 95 8570 0 21 50 6090 21 0 48 5395 50 48 0 2185 60 53 21 0;Global optimal solution found.Objective value: 245.0000Extended solver steps: 0Total solver iterations: 84Variable Value Reduced Cost N 5.000000 0.000000 U( 1) 0.000000 0.000000 U( 2) 0.000000 0.000000 U( 3) 1.000000 0.000000 U( 4) 2.000000 0.000000 U( 5) 3.000000 0.000000 DIST( 1, 1) 0.000000 0.000000DIST( 1, 2) 70.00000 0.000000 DIST( 1, 3) 90.00000 0.000000 DIST( 1, 4) 95.00000 0.000000 DIST( 1, 5) 85.00000 0.000000 DIST( 2, 1) 70.00000 0.000000 DIST( 2, 2) 0.000000 0.000000 DIST( 2, 3) 21.00000 0.000000 DIST( 2, 4) 50.00000 0.000000 DIST( 2, 5) 60.00000 0.000000 DIST( 3, 1) 90.00000 0.000000 DIST( 3, 2) 21.00000 0.000000 DIST( 3, 3) 0.000000 0.000000 DIST( 3, 4) 48.00000 0.000000 DIST( 3, 5) 53.00000 0.000000 DIST( 4, 1) 95.00000 0.000000 DIST( 4, 2) 50.00000 0.000000 DIST( 4, 3) 48.00000 0.000000 DIST( 4, 4) 0.000000 0.000000 DIST( 4, 5) 21.00000 0.000000 DIST( 5, 1) 85.00000 0.000000 DIST( 5, 2) 60.00000 0.000000 DIST( 5, 3) 53.00000 0.000000 DIST( 5, 4) 21.00000 0.000000 DIST( 5, 5) 0.000000 0.000000 X( 1, 1) 0.000000 0.000000 X( 1, 2) 1.000000 70.00000 X( 1, 3) 0.000000 90.00000 X( 1, 4) 0.000000 95.00000 X( 1, 5) 0.000000 85.00000 X( 2, 1) 0.000000 70.00000 X( 2, 2) 0.000000 0.000000 X( 2, 3) 1.000000 21.00000 X( 2, 4) 0.000000 50.00000 X( 2, 5) 0.000000 60.00000 X( 3, 1) 0.000000 90.00000 X( 3, 2) 0.000000 21.00000 X( 3, 3) 0.000000 0.000000 X( 3, 4) 1.000000 48.00000 X( 3, 5) 0.000000 53.00000 X( 4, 1) 0.000000 95.00000 X( 4, 2) 0.000000 50.00000 X( 4, 3) 0.000000 48.00000 X( 4, 4) 0.000000 0.000000 X( 4, 5) 1.000000 21.00000X( 5, 1) 1.000000 85.00000 X( 5, 2) 0.000000 60.00000 X( 5, 3) 0.000000 53.00000 X( 5, 4) 0.000000 21.00000 X( 5, 5) 0.000000 0.000000Row Slack or Surplus Dual Price1 0.000000 0.0000002 245.0000 -1.0000003 0.000000 0.0000004 0.000000 0.0000005 0.000000 0.0000006 0.000000 0.0000007 0.000000 0.0000008 0.000000 0.0000009 0.000000 0.00000010 0.000000 0.00000011 0.000000 0.00000012 0.000000 0.00000013 0.000000 0.00000014 6.000000 0.00000015 7.000000 0.00000016 3.000000 0.00000017 0.000000 0.00000018 6.000000 0.00000019 2.000000 0.00000020 3.000000 0.00000021 0.000000 0.00000022 1.000000 0.00000023 2.000000 0.00000024 3.000000 0.00000025 3.000000 0.00000026 2.000000 0.00000027 1.000000 0.00000028 0.000000 0.000000 PBCDE P-E-D-B-C-Pdist=0 115 90 95 85115 0 30 32 5590 30 0 48 5395 32 48 0 2185 55 53 21 0;Global optimal solution found.Objective value: 258.0000Extended solver steps: 0Total solver iterations: 12Variable Value Reduced Cost N 5.000000 0.000000 U( 1) 0.000000 0.000000 U( 2) 2.000000 0.000000 U( 3) 3.000000 0.000000 U( 4) 1.000000 0.000000 U( 5) 0.000000 0.000000 DIST( 1, 1) 0.000000 0.000000 DIST( 1, 2) 115.0000 0.000000 DIST( 1, 3) 90.00000 0.000000 DIST( 1, 4) 95.00000 0.000000 DIST( 1, 5) 85.00000 0.000000 DIST( 2, 1) 115.0000 0.000000 DIST( 2, 2) 0.000000 0.000000DIST( 2, 3) 30.00000 0.000000 DIST( 2, 4) 32.00000 0.000000 DIST( 2, 5) 55.00000 0.000000 DIST( 3, 1) 90.00000 0.000000 DIST( 3, 2) 30.00000 0.000000 DIST( 3, 3) 0.000000 0.000000 DIST( 3, 4) 48.00000 0.000000 DIST( 3, 5) 53.00000 0.000000 DIST( 4, 1) 95.00000 0.000000 DIST( 4, 2) 32.00000 0.000000 DIST( 4, 3) 48.00000 0.000000 DIST( 4, 4) 0.000000 0.000000 DIST( 4, 5) 21.00000 0.000000 DIST( 5, 1) 85.00000 0.000000 DIST( 5, 2) 55.00000 0.000000 DIST( 5, 3) 53.00000 0.000000 DIST( 5, 4) 21.00000 0.000000 DIST( 5, 5) 0.000000 0.000000 X( 1, 1) 0.000000 0.000000 X( 1, 2) 0.000000 115.0000 X( 1, 3) 0.000000 90.00000 X( 1, 4) 0.000000 95.00000 X( 1, 5) 1.000000 85.00000 X( 2, 1) 0.000000 115.0000 X( 2, 2) 0.000000 0.000000 X( 2, 3) 1.000000 30.00000 X( 2, 4) 0.000000 32.00000 X( 2, 5) 0.000000 55.00000 X( 3, 1) 1.000000 90.00000 X( 3, 2) 0.000000 30.00000 X( 3, 3) 0.000000 0.000000 X( 3, 4) 0.000000 48.00000 X( 3, 5) 0.000000 53.00000 X( 4, 1) 0.000000 95.00000 X( 4, 2) 1.000000 32.00000 X( 4, 3) 0.000000 48.00000 X( 4, 4) 0.000000 0.000000 X( 4, 5) 0.000000 21.00000 X( 5, 1) 0.000000 85.00000 X( 5, 2) 0.000000 55.00000 X( 5, 3) 0.000000 53.00000 X( 5, 4) 1.000000 21.00000 X( 5, 5) 0.000000 0.000000Row Slack or Surplus Dual Price1 0.000000 0.0000002 258.0000 -1.0000003 0.000000 0.0000004 0.000000 0.0000005 0.000000 0.0000006 0.000000 0.0000007 0.000000 0.0000008 0.000000 0.0000009 0.000000 0.00000010 0.000000 0.00000011 0.000000 0.00000012 0.000000 0.00000013 0.000000 0.00000014 3.000000 0.00000015 2.000000 0.00000016 3.000000 0.00000017 2.000000 0.00000018 1.000000 0.00000019 0.000000 0.00000020 6.000000 0.00000021 3.000000 0.00000022 6.000000 0.00000023 7.000000 0.00000024 0.000000 0.00000025 1.000000 0.00000026 0.000000 0.00000027 2.000000 0.00000028 3.000000 0.000000model:sets:city / 1.. 5/: u;link( city, city):dist, ! 距离矩阵;x;endsetsn = @size( city);data: !距离矩阵，它并不需要是对称的;dist=0 70 115 90 9570 0 46 21 50115 46 0 30 3290 21 30 0 4895 50 32 48 0;enddata!目标函数;min = @sum( link: dist * x);@FOR( city( K):!进入城市K;@sum( city( I)| I #ne# K: x( I, K)) = 1;!离开城市K;@sum( city( J)| J #ne# K: x( K, J)) = 1;);!保证不出现子圈;@for(city(I)|I #gt# 1:@for( city( J)| J#gt#1 #and# I #ne# J:u(I)-u(J)+n*x(I,J)<=n-1););!限制u的范围以加速模型的求解，保证所加限制并不排除掉TSP问题的最优解; @for(city(I) | I #gt# 1: u(I)<=n-2 );!定义X为0\1变量;@for( link: @bin( x));EndGlobal optimal solution found.Objective value: 900.0000Objective bound: 900.0000 Infeasibilities: 0.000000Extended solver steps: 1Total solver iterations: 357Variable Value Reduced CostX( 1) 0.000000 0.000000X( 2) 0.000000 0.000000X( 3) 0.000000 1.000000X( 4) 10.00000 0.000000X( 5) 0.000000 1.000000X( 6) 240.0000 0.000000X( 7) 200.0000 0.000000X( 8) 0.000000 1.000000X( 9) 240.0000 0.000000X( 10) 210.0000 0.000000X( 11) 210.0000 0.000000R( 1, 1) 0.000000 0.000000R( 1, 2) 0.000000 0.000000R( 2, 1) 1.000000 0.000000R( 2, 2) 0.000000 0.000000 R( 3, 1) 0.000000 0.000000 R( 3, 2) 1.000000 0.000000 R( 4, 1) 0.000000 0.000000 R( 4, 2) 1.000000 0.000000 R( 5, 1) 1.000000 0.000000 R( 5, 2) 0.000000 0.000000 R( 6, 1) 1.000000 0.000000 R( 6, 2) 0.000000 0.000000 R( 7, 1) 0.000000 0.000000 R( 7, 2) 1.000000 0.000000 R( 8, 1) 1.000000 0.000000 R( 8, 2) 0.000000 0.000000 R( 9, 1) 0.000000 0.000000 R( 9, 2) 1.000000 0.000000 R( 10, 1) 1.000000 0.000000 R( 10, 2) 0.000000 0.000000 R( 11, 1) 0.000000 0.000000 R( 11, 2) 0.000000 0.000000Row Slack or Surplus Dual Price1 900.0000 1.0000002 0.000000 0.0000003 0.000000 0.0000004 0.000000 0.0000005 0.000000 0.0000006 0.000000 0.0000007 0.000000 0.0000008 0.000000 0.0000009 0.000000 0.00000010 0.000000 0.00000011 0.000000 0.00000012 0.000000 0.00000013 0.000000 0.00000014 0.000000 0.00000015 0.000000 0.00000016 0.000000 0.00000017 0.000000 0.00000018 0.000000 0.00000019 0.000000 0.00000020 0.000000 0.00000021 0.000000 0.00000022 0.000000 0.00000023 0.000000 0.00000024 0.000000 0.00000025 0.000000 0.00000026 0.000000 0.00000027 0.000000 0.00000028 0.000000 0.00000029 0.000000 0.00000030 0.000000 0.00000031 0.000000 0.00000032 0.000000 0.00000033 0.000000 0.00000034 0.000000 0.00000035 0.000000 0.00000036 0.000000 0.00000037 0.000000 0.00000038 0.000000 0.00000039 0.000000 0.00000040 0.000000 0.00000041 0.000000 0.00000042 0.000000 0.00000043 0.000000 0.00000044 0.000000 0.00000045 0.000000 0.00000046 6.000000 0.00000047 6.000000 0.00000048 5.000000 0.00000049 6.000000 0.00000050 6.000000 0.00000051 5.000000 0.00000052 6.000000 0.00000053 5.000000 0.00000054 5.000000 0.00000055 6.000000 0.00000056 5.000000 0.00000057 6.000000 0.00000058 6.000000 0.00000059 5.000000 0.00000060 5.000000 0.00000061 6.000000 0.00000062 6.000000 0.00000063 5.000000 0.00000064 5.000000 0.00000065 6.000000 0.00000066 6.000000 0.00000067 6.000000 0.00000068 0.000000 1.00000069 0.000000 1.00000070 250.0000 0.00000071 470.0000 0.00000072 250.0000 0.00000073 0.000000 0.00000074 40.00000 0.00000075 70.00000 0.00000076 0.000000 1.00000077 0.000000 1.00000078 280.0000 0.00000079 190.0000 0.000000。

旅行商问题的python代码

旅行商问题的python代码旅行商问题 (TSP)旅行商问题是一种经典的组合优化问题，涉及寻找一组城市中的最短环路，每座城市只访问一次并返回起点。

贪心算法一种解决 TSP 的贪心算法是最近邻算法。

该算法从任意城市开始，选择与当前城市距离最近的未访问城市，以此类推，直到访问所有城市并返回起点。

Python 代码以下是使用最近邻算法解决 TSP 的 Python 代码：```pythonimport mathdef distance(city1, city2):"""计算两个城市之间的距离。

"""x1, y1 = city1x2, y2 = city2return math.sqrt((x2 - x1) 2 + (y2 - y1) 2) def tsp(cities):"""使用最近邻算法解决 TSP。

"""current_city = cities[0]visited = set([current_city])path = [current_city]while len(visited) < len(cities):next_city = Nonemin_distance = float('inf')for city in cities:if city not in visited anddistance(current_city, city) < min_distance:next_city = citymin_distance = distance(current_city, city)visited.add(next_city)path.append(next_city)current_city = next_citypath.append(cities[0]) # 返回起点return path# 示例城市cities = [(0, 0), (10, 0), (0, 10), (10, 10)]# 求解最短环路path = tsp(cities)# 打印最短环路print(path)```动态规划另一种解决 TSP 的方法是动态规划。

Java实现旅行商问题

Java实现旅⾏商问题1 问题描述何为旅⾏商问题？按照⾮专业的说法，这个问题要求找出⼀条n个给定的城市间的最短路径，使我们在回到触发的城市之前，对每个城市都只访问⼀次。

这样该问题就可以表述为求⼀个图的最短哈密顿回路的问题。

（哈密顿回路：定义为⼀个对图的每个顶点都只穿越⼀次的回路）很容易看出来，哈密顿回路也可以定义为n+1个相邻顶点v1,v2,v3,…,vn,v1的⼀个序列。

其中，序列的第⼀个顶点和最后⼀个顶点是相同的，⽽其它n-1个顶点都是互不相同的。

并且，在不失⼀般性的前提下，可以假设，所有的回路都开始和结束于相同的特定顶点。

因此，可以通过⽣成n-1个中间城市的组合来得到所有的旅⾏线路，计算这些线路的长度，然后求取最短的线路。

下图是该问题的⼀个⼩规模实例，并⽤该⽅法得到了它的解，2 解决⽅案 2.1 蛮⼒法此处使⽤蛮⼒法解决旅⾏商问题，取的是4个城市规模，并已经定义好各个城市之间的距离（PS：该距离使⽤⼆维数组初始化定义，此处的距离是根据图1中所⽰距离定义）。

此处主要是在体验使⽤蛮⼒法解决该问题的思想，如要丰富成普遍规模问题，还请⼤家⾃⼰稍微修改⼀下哒。

对于代码中如碰到不能理解的地⽅，可以参考⽂章末尾给出的参考资料链接，以及相关代码注解~具体代码如下：package com.liuzhen.chapterThree;public class TravelingSalesman {public int count = 0; //定义全局变量，⽤于计算当前已⾏⾛⽅案次数，初始化为0public int MinDistance = 100; //定义完成⼀个⾏⾛⽅案的最短距离，初始化为100（PS：100此处表⽰⽐实际要⼤很多的距离）public int[][] distance = {{0,2,5,7},{2,0,8,3},{5,8,0,1},{7,3,1,0}}; //使⽤⼆维数组的那个⾳图的路径相关距离长度/** start为开始进⾏排序的位置* step为当前正在⾏⾛的位置* n为需要排序的总位置数* Max为n!值*/public void Arrange(int[] A,int start,int step,int n,int Max){if(step == n){ // 当正在⾏⾛的位置等于城市总个数时++count; //每完成⼀次⾏⾛⽅案，count⾃增1printArray(A); //输出⾏⾛路线⽅案及其总距离}if(count == Max)System.out.println("已完成全部⾏⾛⽅案,最短路径距离为:"+MinDistance);else{for(int i = start;i < n;i++){/*第i个数分别与它后⾯的数字交换就能得到新的排列,从⽽能够得到n!次不同排序⽅案* （PS：此处代码中递归的执⾏顺序有点抽象，具体解释详见本⼈另⼀篇博客：）*算法笔记_017:递归执⾏顺序的探讨（Java）*/swapArray(A,start,i);Arrange(A,start+1,step+1,n,Max);swapArray(A,i,start);}}}//交换数组中两个位置上的数值public void swapArray(int[] A,int p,int q){int temp = A[p];A[p] = A[q];A[q] = temp;}//输出数组A的序列，并输出当前⾏⾛序列所花距离，并得到已完成的⾏⾛⽅案中最短距离public void printArray(int[] A){for(int i = 0;i < A.length;i++) //输出当前⾏⾛⽅案的序列System.out.print(A[i]+" ");int tempDistance = distance[A[0]][A[3]]; //此处是因为，最终要返回出发地城市，所以总距离要加上最后到达的城市到出发点城市的距离for(int i = 0;i < (A.length-1);i++) //输出当前⾏⾛⽅案所花距离tempDistance += distance[A[i]][A[i+1]];if(MinDistance > tempDistance) //返回当前已完成⽅案的最短⾏⾛距离MinDistance = tempDistance;System.out.println(" ⾏⾛路程总和："+tempDistance);}public static void main(String[] args){int[] A = {0,1,2,3};TravelingSalesman test = new TravelingSalesman();test.Arrange(A,0,0,4,24); //此处Max = 4!=24}}运⾏结果：1 2 3 ⾏⾛路程总和：181 32 ⾏⾛路程总和：112 13 ⾏⾛路程总和：232 3 1 ⾏⾛路程总和：113 2 1 ⾏⾛路程总和：183 1 2 ⾏⾛路程总和：230 2 3 ⾏⾛路程总和：110 3 2 ⾏⾛路程总和：182 03 ⾏⾛路程总和：232 3 0 ⾏⾛路程总和：183 2 0 ⾏⾛路程总和：113 0 2 ⾏⾛路程总和：231 0 3 ⾏⾛路程总和：181 3 0 ⾏⾛路程总和：230 1 3 ⾏⾛路程总和：110 3 1 ⾏⾛路程总和：233 0 1 ⾏⾛路程总和：183 1 0 ⾏⾛路程总和：111 2 0 ⾏⾛路程总和：231 02 ⾏⾛路程总和：112 1 0 ⾏⾛路程总和：182 0 1 ⾏⾛路程总和：110 2 1 ⾏⾛路程总和：230 1 2 ⾏⾛路程总和：18已完成全部⾏⾛⽅案,最短路径距离为:112.2 减治法旅⾏商问题的核⼼，就是求n个不同城市的全排列，通俗⼀点的话，就是求1n的全排列。

旅行商问题代码

if( n > size ){
cout << "访问越界\n";
return NULL;
}
else{
for( int i = 0; i < n ;++ i ) it = it->nextN;
return it;
}
}
void AdjList::travelbyG( ){//广度遍历
int getsize(){ return size; }//返回size
headNode* operator[]( int );
void destroyNode( headNode* );//删除
void pushNode( int ,int);//创建结点
void pushEage( int, int,int);//创建边结点
value=adf[0];
if( value == 0||value == -1 ){ //创建的节点个数为0，退出。
cout << "没有建立图，程序即将退出";
return ;
}
else{
size = value;//创建节点不为0。
for( int i = 0; i < m; ++ i ) {visited[i] = 0,a[i]=0;} // 0 是未被访问
}
headNode* AdjList::operator[]( int n ){//operator函数，指（竖列）头结点
headNode *it = head;
eageNode *ieage = ihead->next;

旅行商问题用cw节约算法求解的伪代码

旅行商问题用cw节约算法求解的伪代码一、问题描述旅行商问题（TravellingSalesmanProblem,TSP）是一个经典的组合优化问题，涉及到组合数学、运筹学等多个领域。

具体来说，给定一组城市和每对城市之间的距离，求解旅行商完成所有城市访问后返回原出发城市的最短路径。

该问题具有组合特性，求解困难且需要找到一种有效的方法进行求解。

二、CW节约算法CW节约算法是一种基于节约思想的TSP求解算法，通过迭代的方式逐步更新路径，以寻找最短路径。

该算法的主要思想是每次选择当前路径上的城市中距离总和最小的城市，并从该城市出发选择下一个距离总和最小的城市，以此类推，直到遍历完所有城市。

通过不断迭代，算法能够逐渐逼近最优解。

三、伪代码```pseudocode定义函数CW_Saving_Algorithm(distances,cities):初始化一个空列表path和一个变量total_cost=0//遍历所有城市对于i从0到n-1://选择当前路径上的城市中距离总和最小的城市选择一个城市cur_city=min(path[i+1:],key=distances[cur_city,:])[0] //从该城市出发选择下一个距离总和最小的城市选择一个城市next_city=min(path[i+1:],key=distances[:,cur_city])[0] //将选择的城市加入路径中path.append(cur_city)path.append(next_city)//更新总成本total_cost+=distances[cur_city,next_city]//返回路径和总成本返回path,total_cost```四、算法流程1.初始化路径为空列表`path`和总成本为零`total_cost`。

2.遍历所有城市，选择当前路径上的城市中距离总和最小的城市作为当前城市`cur_city`。

3.从当前城市出发，选择下一个距离总和最小的城市`next_city`，并将该城市加入路径中。

动态规划解决旅行商问题附代码x

d[2][{1}]=24}
这样一共循环 (2^(N-1)-1)*(N-1) 次,就把时间复杂度缩小到 O(N*2 N )的级别 . 核心伪代码如下 :
{ for (i =1;i<n;i++)
d[i][0]=c[i][0];
// 初始化第 0 列
for( j=1;j<2^(N-1)-1;j++)
for(i=1 ; i<n ;i++)
------------word 文档可编辑 -------------
#include <ctime> #include <algorithm> using namespace std;
int N;// 节点数 int matr[20][20];// 存邻接矩阵 int d[20][40000]={0};// 存动态填表数据
d(k,{}) = c ki (k ≠, i)找的是最后一个点到 0 点的距离 .递归求解 1 之后 ,再继续求 V’ 之中剩下的点 ,最后找出 min.
如果按照这个思想直接做 ,对于每一个 i 都要递归剩下的 V 中所有的点 ,所以
这样的时间复杂度就近似于 N!,其中有很多重复的工作 .
可以从小的集合到大的集合算 ,并存入一个二维数组 ,这样当加入一个节点时,就可以用到之前的结果 ,如四个点的情况 :
//2^n 次方解法 for (i =1;i<N;i++)
d[i][0]=matr[i][0];
//初始化第 0 列
for(j=1;j<pow(2,N-1)-1;j=getnextj(j)) for(i=1; i<N ;i++) { if(!(j & ( 1<<(i-1) )))// 如果集合 j 中不包含 i { //对 V[j] 中的每个元素 k，计算 d[i][j] = min{matr[i][k] + d[k][{j-k}]}; int jlist[20]={0}; // 存放 j 的二进制数组 int tmpres[20]={0}; int c=0,k=j,l; while(k) { if(k%2) { jlist[c++]=1; }else { jlist[c++]=0; } k/=2; } c=0; for(l=0;l<N;l++) { if(jlist[l]) { tmpres[c++]=matr[i][l+1] + d[l+1][j-(1<<l)]; } } d[i][j] = getmin(tmpres);

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%粒子群算法求解旅行商问题% By lReijclose all;clear all;PopSize=500;%种群大小CityNum = 14;%城市数OldBestFitness=0;%旧的最优适应度值Iteration=0;%迭代次数MaxIteration =2000;%最大迭代次数IsStop=0;%程序停止标志Num=0;%取得相同适应度值的迭代次数c1=0.5;%认知系数c2=0.7;%社会学习系数w=0.96-Iteration/MaxIteration;%惯性系数,随迭代次数增加而递减%节点坐标node=[16.47 96.10; 16.47 94.44; 20.09 92.54; 22.39 93.37; 25.23 97.24;...22.00 96.05; 20.47 97.02; 17.20 96.29; 16.30 97.38; 14.05 98.12;...16.53 97.38; 21.52 95.59; 19.41 97.13; 20.09 94.55];%初始化各粒子，即产生路径种群Group=ones(CityNum,PopSize);for i=1:PopSizeGroup(:,i)=randperm(CityNum)';endGroup=Arrange(Group);%初始化粒子速度（即交换序）Velocity =zeros(CityNum,PopSize);for i=1:PopSizeVelocity(:,i)=round(rand(1,CityNum)'*CityNum); %round取整end%计算每个城市之间的距离CityBetweenDistance=zeros(CityNum,CityNum);for i=1:CityNumfor j=1:CityNumCityBetweenDistance(i,j)=sqrt((node(i,1)-node(j,1))^2+(node(i,2)-node(j,2))^2); endend%计算每条路径的距离for i=1:PopSizeEachPathDis(i) = PathDistance(Group(:,i)',CityBetweenDistance);endIndivdualBest=Group;%记录各粒子的个体极值点位置,即个体找到的最短路径IndivdualBestFitness=EachPathDis;%记录最佳适应度值,即个体找到的最短路径的长度[GlobalBestFitness,index]=min(EachPathDis);%找出全局最优值和相应序号%初始随机解figure;subplot(2,2,1);PathPlot(node,CityNum,index,IndivdualBest);title('随机解');%寻优while(IsStop == 0) & (Iteration < MaxIteration)%迭代次数递增Iteration = Iteration +1;%更新全局极值点位置,这里指路径for i=1:PopSizeGlobalBest(:,i) = Group(:,index);end%求pij-xij ,pgj-xij交换序，并以概率c1，c2的保留交换序pij_xij=GenerateChangeNums(Group,IndivdualBest);pij_xij=HoldByOdds(pij_xij,c1);pgj_xij=GenerateChangeNums(Group,GlobalBest);pgj_xij=HoldByOdds(pgj_xij,c2);%以概率w保留上一代交换序Velocity=HoldByOdds(Velocity,w);Group = PathExchange(Group,Velocity); %根据交换序进行路径交换Group = PathExchange(Group,pij_xij);Group = PathExchange(Group,pgj_xij);for i = 1:PopSize % 更新各路径总距离EachPathDis(i) = PathDistance(Group(:,i)',CityBetweenDistance);endIsChange = EachPathDis<IndivdualBestFitness;%更新后的距离优于更新前的，记录序号 IndivdualBest(:, find(IsChange)) = Group(:, find(IsChange));%更新个体最佳路径IndivdualBestFitness = IndivdualBestFitness.*( ~IsChange) + EachPathDis.*IsChange;%更新个体最佳路径距离[GlobalBestFitness, index] = min(EachPathDis);%更新全局最佳路径,记录相应的序号if GlobalBestFitness==OldBestFitness %比较更新前和更新后的适应度值;Num=Num+1; %相等时记录加一;elseOldBestFitness=GlobalBestFitness;%不相等时更新适应度值，并记录清零;Num=0;endif Num >= 20 %多次迭代的适应度值相近时程序停止IsStop=1;endBestFitness(Iteration) =GlobalBestFitness;%每一代的最优适应度end%最优解subplot(2,2,2);PathPlot(node,CityNum,index,IndivdualBest);title('优化解');%进化曲线subplot(2,2,3);plot((1:Iteration),BestFitness(1:Iteration));grid on;title('进化曲线');%最小路径值GlobalBestFitnessfunction Group=Arrange(Group)[x y]=size(Group);[NO1,index]=min(Group',[],2); %找到最小值1for i=1:ypop=Group(:,i);temp1=pop([1: index(i)-1]);temp2=pop([index(i): x]);Group(:,i)=[temp2' temp1']';endfunction ChangeNums=GenerateChangeNums(Group,BestVar);[x y]=size(Group);ChangeNums=zeros(x,y);for i=1:ypop=BestVar(:,i);%从BestVar取出一个顺序pop1=Group(:,i);%从粒子群中取出对应的顺序for j=1:x %从BestVar的顺序中取出一个序号NoFromBestVar=pop(j);for k=1:x %从对应的粒子顺序中取出一个序号NoFromGroup=pop1(k);if (NoFromBestVar==NoFromGroup) && (j~=k) %两序号同且不在同一位置 ChangeNums(j,i)=k; %交换子pop1(k)=pop1(j);pop1(j)=NoFromGroup;endendendendfunction Hold=HoldByOdds(Hold,Odds)[x,y]=size(Hold);for i=1:xfor j=1:yif rand>OddsHold(i,j)=0;endendendfunction SumDistance=PathDistance(path,CityBetweenDistance)L=length(path); %path为一个循环的节点顺序SumDistance=0;for i=1:L-1SumDistance=SumDistance+CityBetweenDistance(path(i),path(i+1));endSumDistance=SumDistance+CityBetweenDistance(path(1),path(L)); %加上首尾节点的距离function Group=PathExchange(Group,Index)[x y]=size(Group);for i=1:ya=Index(:,i); %取出其中一组交换序pop=Group(:,i); %取出对应的粒子for j=1:x %取出其中一个交换算子作交换if a(j)~=0pop1=pop(j);pop(j)=pop(a(j));pop(a(j))=pop1;endendGroup(:,i)=pop;endfunction PathPlot(node,CityNum,index,EachBest);for i=1:CityNumNowBest(i,:)=node((EachBest(i,index)),:);endNowBest(CityNum+1,:)=NowBest(1,:);plot(node(:,1),node(:,2),'*');line(NowBest(:,1),NowBest(:,2));grid on;。