动态规划算法实验报告

实验标题1、矩阵连乘2、最长公共子序列3、最大子段和4、凸多边形最优三角剖分5、流水作业调度6、0-1背包问题7、最优二叉搜索树实验目的掌握动态规划法的基本思想和算法设计的基本步骤。

实验内容与源码1、矩阵连乘#include<iostream>#include<cstdlib>using namespace std;const int size=4;//ra,ca和rb,cb分别表示矩阵A和B的行数和列数void matriMultiply(int a[][4],int b[][4],int c[][4],int ra ,int ca,int rb ,int cb ){if(ca!=rb) cerr<<"矩阵不可乘";for(int i=0;i<ra;i++)for(int j=0;j<cb;j++){int sum=a[i][0]*b[0][j];for(int k=1;k<ca;k++)sum+=a[i][k]*b[k][j];c[i][j]=sum;}}void MatrixChain(int *p,int n,int m[][4],int s[][4]){for(int i=1;i<=n;i++) m[i][i]=0;//对角线for(int r=2;r<=n;r++)//外维for(int i=1;i<=n-r+1;i++)//上三角{int j=i+r-1;m[i][j]=m[i+1][j]+p[i-1]*p[i]*p[j];s[i][j]=i;for(int k=i+1;k<j;k++){int t=m[i][k]+m[k+1][j]+p[i-1]*p[k]*p[j];if(t<m[i][j]){m[i][j]=t;s[i][j]=k;}}}}void Traceback(int i,int j,int s[][4]){if(i == j){cout<<"A"<<i;}else if(i+1 == j){cout<<"(A"<<i<<"A"<<j<<")";}else{cout<<"(";Traceback(i,s[i][j],s);Traceback(s[i][j]+1,j,s);cout<<")";}}int main(){int w;cout<<"矩阵个数：";cin>>w;int p[w],s[w][w];cout<<"输入矩阵A1维数：";cin>>p[0]>>p[1];for(int i=2 ; i<=w ; i++){int m = p[i-1];cout<<"输入矩阵A"<<i<<"维数：";cin>>p[i-1]>>p[i];if(p[i-1] != m){cout<<endl<<"维数不对，矩阵不可乘！"<<endl;exit(1);}}Traceback(1,w,s);return 0;}运行结果2、最长公共子序列#include<cstring>#include<iostream>#define N 100using namespace std;//str1存储字符串x，str2存储字符串ychar str1[N],str2[N];//lcs存储最长公共子序列char lcs[N];//c[i][j]存储str1[1...i]与str2[1...j]的最长公共子序列的长度int c[N][N];//flag[i][j]==0为str1[i]==str2[j]//flag[i][j]==1为c[i-1][j]>=s[i][j-1]//flag[i][j]==-1为c[i-1][j]<s[i][j-1]int flag[N][N];//求长度int LCSLength(char *x, char *y){int i,j;//分别取得x,y的长度int m = strlen(x);int n = strlen(y);for(i=1;i<=m;i++)c[i][0] = 0;for(i=0;i<=n;i++)c[0][i] = 0;for(i=1;i<=m;i++)for(j=1;j<=n;j++){if(x[i-1]==y[j-1]){c[i][j] = c[i-1][j-1] +1;flag[i][j] = 0;}else if(c[i-1][j]>=c[i][j-1]){c[i][j] = c[i-1][j];flag[i][j] = 1;}else{c[i][j] = c[i][j-1];flag[i][j] = -1;}}return c[m][n];}//求出最长公共子序列char* getLCS(char *x, char *y,int len,char *lcs) {int i = strlen(x);int j = strlen(y);while(i&&j){if(flag[i][j]==0){lcs[--len] = x[i-1];i--;j--;}else if(flag[i][j]==1)i--;elsej--;}return lcs;}int main(){int i;cout<<"请输入字符串x："<<endl;cin>>str1;cout<<"请输入字符串y："<<endl;cin>>str2;int lcsLen = LCSLength(str1,str2);cout<<"最长公共子序列长度："<<lcsLen<<endl;char *p = getLCS(str1,str2,lcsLen,lcs);cout<<"最长公共子序列为：";for(i=0;i<lcsLen;i++)cout<<lcs[i]<<" ";return 0;}运行结果3、最大子段和//分治法求最大子段和#include<iostream>using namespace std;int MaxSubSum(int *a,int left,int right){int sum=0;if(left==right) sum=a[left]>0?a[left]:0;else{int center = (left+right)/2;//最大子段和在左边int leftsum=MaxSubSum(a,left,center);//最大子段和在右边int rightsum=MaxSubSum(a,center+1,right);//最大子段和在中间int s1=0;int lefts=0;for(int i=center;i>=left;i--){lefts+=a[i];if(lefts>s1) s1=lefts;}int s2=0;int rights=0;for(int i=center+1;i<=right;i++){rights+=a[i];if(rights>s2) s2=rights;}sum=s1+s2;//前后子段和相加//判断最大子段和if(sum>leftsum)sum=leftsum;if(sum>rightsum) sum=rightsum;}return sum;}int MaxSum(int *a,int n){return MaxSubSum(a,1,n-1);}int main(){int a[8]={2,-3,-5,4,1,7,1,-5};cout<<"最大子段和为："<<MaxSum(a,8);return 0;}//动态规划法#include<iostream>using namespace std;int MaxSum(int *a,int n){int sum=0,b=0;for(int i=1;i<n;i++)//此处不能=n，{if(b>0) b+=a[i];else b=a[i];if(b>sum) sum=b;}return sum;}int main(){int a[8]={2,-3,-5,4,1,7,1,-5};cout<<"最大子段和为："<<MaxSum(a,8);return 0;}运行结果4、凸多边形最优三角剖分#include<iostream>#include<cmath>#include<cstdlib>#define N 50using namespace std;struct point{int x;int y;};int distance(point X, point Y)//两点距离{int dis = (Y.x-X.x)*(Y.x-X.x) + (Y.y-X.y)*(Y.y-X.y);return (int)sqrt(dis);}int w(point a, point b, point c)//权值{return distance(a,b) + distance(b,c) + distance(a,c);}bool JudgeInput()//判断是否能构成凸多边形{point *v; //记录凸多边形各顶点坐标int *total; //记录坐标在直线方程中的值int m,a,b,c;cout<<"请输入凸多边形顶点个数：";cin>>m;int M = m-1;for(int i=0 ; i<m ; i++){cout<<"输入顶点v"<<i<<"的坐标：";cin>>v[i].x>>v[i].y;}//根据顶点坐标判断是否能构成一个凸多边形for(int j=0 ; j<m ; j++){int p = 0;int q = 0;if(m-1 == j){a = v[m-1].y - v[0].y;b = v[m-1].x - v[0].y;c = b * v[m-1].y - a * v[m-1].x;}else{a = v[j].y - v[j+1].y;b = v[j].x - v[j+1].x;c = b * v[j].y - a * v[j].x;}for(int k=0 ; k<m ; k++){total[k] = a * v[k].x - b * v[k].y + c;if(total[k] > 0){p = p+1;}else if(total[k] < 0){q = q+1;}}if((p>0 && q>0) || (p==0 && q==0)){cout<<"无法构成凸多边形！"<<endl;exit(1);}}}bool minWeightTriangulation()//计算最优值算法{int M;int **t, **s;point *v;for(int i=1 ; i<=M ; i++)t[i][i] = 0;for(int r=2 ; r<=M ; r++)for(int i=1 ; i<=M-r+1 ; i++){int j = i+r-1;t[i][j] = t[i+1][j] + w(v[i-1],v[i],v[j]);s[i][j] = i;for(int k=i+1 ; k<i+r-1 ; k++){int u = t[i][k] + t[k+1][j] + w(v[i-1],v[k],v[j]);if(u < t[i][j]){t[i][j] = u;s[i][j] = k;}}}return true;}void Traceback(int i, int j, int **s){if(i == j)return;Traceback(i,s[i][j],s);Traceback(s[i][j]+1,j,s);cout<<"三角形：v"<<i-1<<"v"<<s[i][j]<<"v"<<j<<endl;}int main(){int **s; //记录最优三角剖分中所有三角形信息int **t; //记录最优三角剖分所对应的权函数值point *v; //记录凸多边形各顶点坐标int *total; //记录坐标在直线方程中的值int M=0;t = new int *[N];s = new int *[N];for(int i=0 ; i<N ; i++){t[i] = new int[N];s[i] = new int[N];}v = new point[N];total = new int[N];if(JudgeInput()){if(minWeightTriangulation()){Traceback(1,M,s);cout<<endl;cout<<"最优权值之和为："<<t[1][M]<<endl;}}return 0;}运行结果：5、流水作业调度#include<iostream>#define N 100using namespace std;class Jobtype{public:/* int operator<=(Jobtype a)const{return(key<=a.key);}*/int key;int index;bool job;};void sort(Jobtype *d,int n){int i,j;Jobtype temp;bool exchange; //交换标志for(i = 0;i < n;i ++){ //最多做n-1趟排序exchange = false; //本趟排序开始前，交换标志应为假for(j = n - 1;j >= i;j --)if(d[j+1].key < d[j].key){temp = d[j+1];d[j+1] = d[j];d[j] = temp;exchange=true; //发生了交换，故将交换标志置为真}if(!exchange) //本趟排序未发生交换，提前终止算法return;}}int FlowShop(int n,int *a,int *b,int *c){Jobtype *d = new Jobtype[n];for(int i=0;i<n;i++)//初始化{d[i].key=a[i]>b[i]?b[i]:a[i];// 执行时间d[i].job=a[i]<=b[i];// 作业组d[i].index=i;//作业序号}sort(d,n);;int j=0;int k=n-1;for(int i=0;i<n;i++)//最优调度{if(d[i].job){c[j++]=d[i].index;}else{c[k--]=d[i].index;}}j=a[c[0]];k=j+b[c[0]];for(int i=1;i<n;i++){j+=a[c[i]];k=j<k?k+b[c[i]]:j+b[c[i]];}delete d;//回收空间return k;//返回调度时间}int main(){int n,*a,*b,*c;cout<<"作业数：";cin>>n;Jobtype *d = new Jobtype[N];a=new int[N];b=new int[N];c=new int[N];cout<<"请输入作业号和时间：";for(int i=0;i<n;i++){cin>>d[i].index>>d[i].key;}cout << endl;int k=FlowShop(n,a,b,c);cout<<"\n调度时间："<<k<<endl;cout<<"最优调度序列:";for (int i = 0; i < n; i++) // 输出最优调度序列{cout << c[i] << " ";}return 0;}运行结果：6、0-1背包问题#include <iostream>#include <iomanip>using namespace std;const int C=10;//容量const int N=5;//个数int max(const int a,const int b){return a>b?a:b;}int min(const int a,const int b){return a<b?a:b;}/*m为记录数组m[i][j]代表在剩有j容量的条件下，从i开始往后的物品中可以取得的最大价值w为重量数组，v为价值数组n为物品个数，c为开始容量则m[1][c]即此背包能剩下的最大价值*/void knapsack(int **m,int n, int c,int *w, int *v) {int jMax = min(w[n]-1,c);//前n-1个物品for(int j=0;j<=jMax;j++)m[n][j]=0;for(int j=w[n];j<=c;j++)m[n][j]=v[n];for(int i=n-1;i>1;i--){jMax=min(w[i]-1,c);for(int j=0;j<=jMax;j++)m[i][j] = m[i+1][j];for(int j=w[i];j<=c;j++)m[i][j] = max(m[i+1][j],m[i+1][j-w[i]]+v[i]);}m[1][c]=m[2][c];if(c>=w[1])m[1][c]=max(m[1][c],m[2][c-w[1]]+v[1]);}//找出最优解，0表示不能装，1表示能装void traceback(int **m,int n,int c,int *x,int *w) {for(int i=1;i<n;i++){if(m[i][c]==m[i+1][c]) x[i]=0;else{x[i]=1;c-=w[i];}}x[n]=(m[n][c]==0)?0:1;}int main(){int *v=new int[N+1];int *w=new int[N+1];int **m=new int* [N+1];int *x=new int [N+1];for(int i=0;i<N+1;i++){m[i]=new int[C+1];}cout<<"输入重量序列,"<<N<<"个"<<endl; for(int i=1;i<=N;i++)cin>>w[i];cout<<"输入价值序列,"<<N<<"个"<<endl; for(int i=1;i<=N;i++)cin>>v[i];knapsack(m,N,C,w,v);traceback(m,N,C,x,w);cout<<"最优值："<<m[1][C]<<endl; cout<<"是否装入背包的情况：";for(int i=1;i<=N;i++){cout<<x[i];}for(int i=0;i<N+1;i++){delete m[i];}delete []m;return 0;}运行结果7、最优二叉搜索树#include<iostream>#include<cmath>#include<limits>#define N 100using namespace std;const double MAX = numeric_limits<double>::max(); //double的最大值//a[i]为结点i被访问的概率//b[i]为“虚结点”i被访问的概率//m[i][j]用来存放子树（i，j）的期望代价//w[i][j]用来存放子树（i，j）的所有结点（包括虚结点）的a,b概率之和//s[i][j]用来跟踪root的void OptimalBinarySearchTree(double *a,double *b,int n){int s[N][N];double m[N][N];double w[N][N];int i,j,l,r;for(i=1; i<=n+1; i++){m[i][i-1] = b[i-1];w[i][i-1] = b[i-1];}for(l=1; l<=n; l++){for(i=1; i<=n-l+1; i++){j = l+i-1;m[i][j] = MAX;w[i][j] = w[i][j-1] + a[j] +b[j];for(r=i; r<=j; r++){double k = m[i][r-1] + w[i][j] + m[r+1][j];if(k<m[i][j]){m[i][j] = k;s[i][j] = k;}}}}cout<<m[1][n];}int main(){double a[N],b[N];int n;double sum = 0;int i,j,l;cout<<"请输入关键字的个数："<<endl;cin>>n;cout<<"请输入每个关键字的概率："<<endl;for(i=1; i<=n; i++){cin>>a[i];sum += a[i];}cout<<"请输入每个虚拟键的概率："<<endl;for(i=0; i<=n; i++){cin>>b[i];sum += b[i];}if(abs(sum-1)>0.01){cout<<"输入的概率和不为1,请重新输入"<<endl;}cout<<"最优二叉查找树的期望搜索代价为：";OptimalBinarySearchTree(a,b,n);return 0;}运行结果：实验总结通过实现动态规划的这个题目，对动态规划算法有了进一步的了解。