矩阵连乘问题

} if(i==j) return0; intu=LookupChain(i,i)+LookupChain(i+1,j)+p[i-1]*p[i]*p[j]; s[i][j]=i; for(intk=i+1;k<j;k++) { intt=LookupChain(i,k)+LookupChain(k+1,j)+p[i-1]*p[k]*p[j]; if(t<u) { u=t; s[i][j]=k; } } m[i][j]=u; returnu;}intMemoizedMatrixChain(intn,int**m,int**s){ for(inti=1;i<=n;i++) for(intj=i;j<=n;j++)m[i][j]=0; //把?m[i][j]的?矩?阵ó的?上?三▂角?全?部?赋3值μ为a0。￡returnLookupChain(1,n);}//动ˉ态?规?划?法ぁ?voidMatrixChain(){ for(inti=1;i<=n;i++) { m[i][i]=0; } for(intr=2;r<=n;r++) { for(inti=1;i<=n-r+1;i++) {intj=r+i-1;m[i][j]=m[i][i]+m[i+1][j]+p[i-1]*p[i]*p[j];s[i][j]=i;for(intk=i+1;k<j;k++) {inttemp=m[i][k]+m[k+1][j]+p[i-1]*p[k]*p[j];if(temp<m[i][j]) {m[i][j]=temp; s[i][j]=k; }} } }}//直±接ó递蘗归é法ぁ?intRecurMatrixChain(inti,intj){ if(i==j){return0;} intu=RecurMatrixChain(i,i)+RecurMatrixChain(i+1,j)+p[i-1]*p[i]*p[j];//递蘗归é，?p[]为a维?数簓 s[i][j]=i;//记?录?加ó括ぁ?号?的?位?置? for(intk=i+1;k<j;k++) { intt=RecurMatrixChain(i,k)+RecurMatrixChain(k+1,j)+p[i-1]*p[k]*p[j];//递蘗归é if(t<u){u=t;s[i][j]=k;}//判D断?哪?个?值μ更ü小?，?选?取?哪?个? } returnu;}voidTraceback(inti,intj){if(i==j)//如?果?只?有瓺一?个?矩?阵ó，?则ò直±接ó输?出?{cout<<"A"<<i;}elseif(i+1==j)//加ó括ぁ?号?输?出?{cout<<"(A"<<i<<"A"<<j<<")";}else{cout<<"(";Traceback(i,s[i][j]);Traceback(s[i][j]+1,j);cout<<")";}}MatrixMultiply.cpp#include<iostream>#include"matrix.h"usingnamespacestd;//intn;//定¨义?一?个?全?局?变?量?intmain(){L: intchoice; cout<<"请输入相乘的矩阵的个数n="; cin>>n; cout<<"请输入第1个矩阵行数和第1个矩阵到第"<<n<<"个矩阵的列数："; for(inti=0;i<=n;i++) { cin>>p[i]; } cout<<endl; cout<<"请选择矩阵连乘的算法:"<<endl; cout<<"1.备忘录算法"<<endl; cout<<"2.直接递归算法"<<endl; cout<<"3.动态规划算法"<<endl; cout<<"4.重新输入矩阵"<<endl; cout<<"5.退出程序"<<endl;cout<<endl; cout<<"请输入选择的编号："; cin>>choice; cout<<endl; while(choice!=5){ switch(choice){ case1:LookupChain(1,n); cout<<"动态规划算法："<<endl; cout<<"矩阵连乘的最优值为："<<m[1][n]<<endl; cout<<"矩阵连乘的最优解为："; Traceback(1,n); cout<<endl; break; case2: RecurMatrixChain(0,n); cout<<"动态规划算法："<<endl; cout<<"矩阵连乘的最优值为："<<m[1][n]<<endl; cout<<"矩阵连乘的最优解为："; Traceback(1,n); cout<<endl; break; case3: MatrixChain(); cout<<"动态规划算法："<<endl; cout<<"矩阵连乘的最优值为："<<m[1][n]<<endl; cout<<"矩阵连乘的最优解为："; Traceback(1,n); cout<<endl; break; case4: gotoL; break; case5: choice=4; break; default: break;} cout<<endl; cout<<"请选择矩阵连乘的算法:"<<endl; cout<<"1.备忘录算法"<<endl; cout<<"2.直接递归算法"<<endl