dp矩阵连乘

#include <iostream>
#include <stdio.h>
using namespace std;
#define NUM 51
int p[NUM];
int m[NUM][NUM];
int s[NUM][NUM];
void MatrixChain(int n)
{
	for (int i = 1; i <= n; i++) m[i][i] = 0;
	for (int r = 2; r <= n; r++)
		for (int i = 1; i <= n - r + 1; i++)
		{
			int j = i + r - 1;
			//计算初值，从i处断开
			m[i][j] = m[i + 1][j] + p[i - 1] * p[i] * p[j];//r=1时可以计算出所有相邻矩阵的乘积
			s[i][j] = i;
			for (int k = i + 1; k<j; k++)//通过循环判断各种组合大小
			{
				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)
{
	if (i == j) printf("A%d", i);
	else
	{
		printf("(");
		TraceBack(i, s[i][j]);//s[i][j]表示A[i]+....A[j]中的最佳分割点。此处分割为
		TraceBack(s[i][j] + 1, j);//分割的后一部分
		printf(")");
	}
}
int main() {
	int x, i = 1;
	while (cin >> x) {
		p[i++] = x;
	}
	MatrixChain(i - 1);
	TraceBack(1, i - 1);
	return 0;
}

简化版的两种

int Recurve(int i, int j)
{
　　if (i == j) return 0;
　　int u = Recurve(i, i)+Recurve(i+1,j)+p[i-1]*p[i]*p[j];
　　s[i][j] = i;
　　for (int k = i+1; k<j; k++) 
　　{
	int t = Recurve(i, k) + Recurve(k+1,j)+p[i-1]*p[k]*p[j];
	if (t<u) { u = t; s[i][j] = k;}
　　}
　　m[i][j] = u;
　　return u;
} 

int LookupChain  (int i, int j)
{
　　if (m[i][j]>0) return m[i][j];
　　if (i==j) return 0;
　　int u = LookupChain(i,i)+LookupChain(i+1,j)+p[i-1]*p[i]*p[j];
　　s[i][j] = i;
　　for (int k = i+1; k<j; k++) 
　　{
	int t = 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;
　　return u;
}