数学建模必备程序代码网站

时间问题
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;
ENDSETS
DATA: !时间矩阵;
T = 0 1.7 1.5 1.4 3 1.9 2.8
1.7 0 3
2.6 4 3
3.5
1.5 3 0 1.5 3.7
2.6 2.8
1.4
2.6 1.5 0 11.5 3
3.2
3 4 3.7 11.5 0 3.6 3.3
1.9 3
2.6 3
3.6 0 15.6
2.8
3.5 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
距离问题
/demo_c92_i146007.html
距离问题
Lingo代码：
MODEL:
SETS:
CITY / 1.. 6/: U; ! U( I) = sequence no. of city;
LINK( CITY, CITY):
DIST, ! The distance matrix;
X; ! X( I, J) = 1 if we use link I, J;
ENDSETS
DATA: !Distance matrix, it need not be symmetric;
DIST************
56 0 21 57 78 70
35 21 0 36 68 68
21 57 36 0 51 61
51 78 68 51 0 13
60 70 68 61 13 0;
ENDDATA
!The model:Ref. Desrochers & Laporte, OR Letters,
Feb. 91;
N = @SIZE( CITY);
MIN = @SUM( LINK: DIST * X);
@FOR( CITY( K):
! It must be entered;
@SUM( CITY( I)| I #NE# K: X( I, K)) = 1;
! It must be departed;
@SUM( CITY( J)| J #NE# K: X( K, J)) = 1;
! Weak form of the subtour breaking constraints;
! These are not very powerful for large problems;
@FOR( CITY( J)| J #GT# 1 #AND# J #NE# K:
U( J) >= U( K) + X ( K, J) -
( N - 2) * ( 1 - X( K, J)) +
( N - 3) * X( J, K)));
! Make the X's 0/1;
@FOR( LINK: @BIN( X));
! For the first and last stop we know...;
@FOR( CITY( K)| K #GT# 1:
U( K) <= N - 1 - ( N - 2) * X( 1, K);
U( K) >= 1 + ( N - 2) * X( K, 1));
END
matlab代码：
function main
clc,clear
global a
% a=zeros(6);
% a(1,2)=56;a(1,3)=35;a(1,4)=21;a(1,5)=51;a(1,6)=60; % a(2,3)=21;a(2,4)=57;a(2,5)=78;a(2,6)=70;
% a(3,4)=36;a(3,5)=68;a(3,6)=68; a(4,5)=51;a(4,6)=61; % a(5,6)=13; a=a+a';
load cost
a=Muti_Cost;%边权矩阵
L=size(a,1);
c1=1:53; %初始圈
[circle,long]=modifycircle(c1,L);
c2=[1 53 2:52];%改变初始圈，该算法的最后一个顶点不动[circle2,long2]=modifycircle(c2,L);
if long2<long
long=long2;
circle=circle2;
end
circle,long
%*******************************************
%修改圈的子函数
%*******************************************
function [circle,long]=modifycircle(c1,L);
global a
flag=1;
while flag>0
flag=0;
for m=1:L-3
for n=m+2:L-1
if a(c1(m),c1(n))+a(c1(m+1),c1(n+1))<...
a(c1(m),c1(m+1))+a(c1(n),c1(n+1))
flag=1;
c1(m+1:n)=c1(n:-1:m+1);
end
end
end
end
long=a(c1(1),c1(L));
for i=1:L-1
long=long+a(c1(i),c1(i+1));
end
circle=c1;
/s/blog_a16714bf0101d7jl.html
TSP的Lingo+matlab求解
旅行商问题，也即称货郎担问题（TSP）的描述为：
有一个推销员，从城市1出发，要遍访城市2，3，…，n各一次，最后返回城市1。

已知从城市i到j的旅费为，问他应按怎样的次序访问这些城市，使得总旅费最少？
TSP已被证明是NP难问题，目前还没有发现多项式时间的算法。

对于小规模问题，我们求解这个混合整数线性规划问题的方式还是有效的。

现有问题：平面里有N个城市的坐标，求最短路径，遍历所有城市，且每个城市只经过一次。

下面利用Lingo+matlab求解。

利用matlab随机生成10个城市的坐标，Lingo读取matlab生成的10个城市的坐标，利用Lingo强大的非线性整数规划的功能算出结果，导出数据供matlab画图。

MODEL:
! Traveling Salesman Problem for the cities of
Atlanta, Chicago, Cincinnati, Houston, LA,
Montreal;
SETS:
CITY / 1.. 6/: U; ! U( I) = sequence no. of city;
LINK( CITY, CITY):
DIST, ! The distance matrix;
X; ! X( I, J) = 1 if we use link I, J;
ENDSETS
DATA: !Distance matrix, it need not be symmetric;
DIST = 0 702 454 842 2396 1196
702 0 324 1093 2136 764
454 324 0 1137 2180 798
842 1093 1137 0 1616 1857
2396 2136 2180 1616 0 2900
1196 764 798 1857 2900 0;
ENDDATA
!The model:Ref. Desrochers & Laporte, OR Letters,
Feb. 91;
N = @SIZE( CITY);
MIN = @SUM( LINK: DIST * X);
@FOR( CITY( K):
! It must be entered;
@SUM( CITY( I)| I #NE# K: X( I, K)) = 1;
! It must be departed;
@SUM( CITY( J)| J #NE# K: X( K, J)) = 1;
! Weak form of the subtour breaking constraints;
! These are not very powerful for large problems;
@FOR( CITY( J)| J #GT# 1 #AND# J #NE# K:
U( J) >= U( K) + X ( K, J) -
( N - 2) * ( 1 - X( K, J)) +
( N - 3) * X( J, K)
);
);
! Make the X's 0/1;
@FOR( LINK: @BIN( X));
! For the first and last stop we know...;
@FOR( CITY( K)| K #GT# 1:
U( K) <= N - 1 - ( N - 2) * X( 1, K);
U( K) >= 1 + ( N - 2) * X( K, 1)
);
END
/link?url=x96rsjznkqm1Wt8mzSBIS3zfjynpF-QwgV6bFA3SbL1uK ROE1oDPzJzF_PgpEaluJUcweejp2fEs3C4-yyDPv_
MODEL:
SETS:
city/A1..A13/:U;!U(i)=cicy No;
links(city,city):distance,!the distance of the matrix;
X;!x(i,j)=1 if we use link i,j;
ENDSETS
DATA: !distance matrix;
!10000,the distance of impossible link;
distance= 0, 62, 92, 73,10000,10000,10000,10000,10000,10000,10000,10000,10000,
62, 0,10000,10000, 58,10000,10000,10000,10000,10000,10000,10000,10000,
92,10000, 0,10000,10000, 52, 57,10000,10000,10000,10000,10000,10000,
73,10000,10000, 0,10000,10000, 59,10000,10000,10000,10000,10000,10000,
10000, 58,10000,10000, 0,10000,10000,104,10000,10000,10000,10000,10000,
10000,10000, 52,10000,10000, 0,10000,10000, 85,104,10000, 77,10000,
10000,10000, 57, 59,10000,10000, 0,10000, 81,10000,10000,10000,10000,
10000,10000,10000,10000,104,10000,10000, 0,10000,115,166,10000,10000,
10000,10000,10000,10000,10000, 85, 81,10000, 0,10000,10000,10000, 63,
10000,10000,10000,10000,10000,104,10000,115,10000, 0,10000,10000,10000,
10000,10000,10000,10000,10000,10000,10000,166,10000,10000, 0, 87, 73,
10000,10000,10000,10000,10000, 77,10000,10000,10000,10000, 87, 0,10000,
10000,10000,10000,10000,10000,10000,10000,10000, 63,10000, 73,10000, 0; ENDDATA
n=@SIZE(city);
@FOR(links:@BIN(x));
MIN=@SUM(links:distance*X);
@FOR(city(k):
@SUM(city(i)|i#ne#k:x(i,k))=1;
@SUM(city(j)|j#ne#k:x(k,j))=1;);
!--------------------------;
!@FOR(city(k):
@FOR(city(j)|j#gt#1 #and# j#ne#k:U(j)>=U(k)+X(k,j)-(N-2)*(1-X(k,j))+(N-3)*X(j,k)); !);
@FOR(city(k):
@FOR(city(j)|j#gt#1 #and# j#ne#k: u(k) - u(j)+ (n-1)*x(k,j) + (n-3)*x(j,k) <= n-2 ); );
!@FOR(city(k)|k#gt#1: U(k)<=N-1-(N-2)*X(1,k); !U(k)>=1+(N-2)*X(k,1)
);
@FOR(city(k)|k#gt#1: X(1,k)+X(k,1)<=1
);
!--------------------------;
!==========================;
!@for(links(i,j)| i#ne#j #and# i#gt#1 : u(i)-u(j)+n*x(i,j)<=n-1 );
!this line has the same function as the lines between "-----", but it needs longer time );
!==========================;
END
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