矩阵连乘算法

//A1 30*35 A2 35*15 A3 15*5 A4 5*10 A5 10*20 A6 20*25
//p[0-6]={30,35,15,5,10,20,25}
#include <stdio.h>
#include <iostream>
using namespace std;

const int L = 7;

int MatrixChain(int n,int **m,int **s,int *p);
void Traceback(int i,int j,int **s);//构造最优解

int main()
{
int p[L]={30,35,15,5,10,20,25};

int **s = new int *[L];
int **m = new int *[L];
for(int i=0;i<L;i++)
{
s[i] = new int[L];
m[i] = new int[L];
}

cout<<"矩阵的最少计算次数为："<<MatrixChain(6,m,s,p)<<endl;
cout<<"矩阵最优计算次序为："<<endl;
Traceback(1,6,s);
return 0;
}

int MatrixChain(int n,int **m,int **s,int *p)
{
for(int i=1; i<=n; i++)
{
m[i][i] = 0;
}
for(int r=2; r<=n; r++) //r为当前计算的链长（子问题规模）
{
for(int i=1; i<=n-r+1; i++)//n-r+1为最后一个r链的前边界
{
int j = i+r-1;//计算前边界为r，链长为r的链的后边界

m[i][j] = m[i+1][j] + p[i-1]*p[i]*p[j];//将链ij划分为A(i) * ( A[i+1:j] )

s[i][j] = i;

for(int k=i+1; k<j; k++)
{
//将链ij划分为( A[i:k] )* (A[k+1:j])
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;
}
}
}
}
return m[1][L-1];
}

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<<"Multiply A"<<i<<","<<s[i][j];
cout<<" and A"<<(s[i][j]+1)<<","<<j<<endl;
}