矩阵连乘问题（动态规划算法）

问题描述：

具体可参考：https://blog.csdn.net/liufeng_king/article/details/8497607

代码如下：

#ifndef MATRIX_CHAIN_H
#define MATRIX_CHAIN_H

void matrix_chain(int *p, int n, int **m, int **s);
void traceback(int i, int j, int **s);

#endif

#include <iostream>
#include "matrix_chain.h"

using namespace std;

//利用动态规划算法获取最优值
void matrix_chain(int *p, int n, int **m, int **s)  //p:各个矩阵的列数，n：矩阵个数，m：m[i:j]矩阵i到j的相乘次数，s:对应的分开位置
{
    for (int i = 0; i < n; i++)
    {
        m[i][i] = 0;
    }
    for (int r = 2; r <= n; r++)
    {
        for (int i = 0; i < n - r + 1; i++)
        {
            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;
                }
            }
        }
    }
}

//利用s[i][j]获取最优解
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;
}

#include <iostream>
#include "matrix_chain.h"

using namespace std;

int main(void)
{
    int matrix_num = 0;  //矩阵个数
    cout << "请输入矩阵个数：" << endl;
    cin >> matrix_num;

    int **m = new int *[matrix_num];
    for (int i = 0; i < matrix_num; i++)
        m[i] = new int[matrix_num];
    int **s = new int *[matrix_num];
    for (int i = 0; i < matrix_num; i++)
        s[i] = new int[matrix_num];
    int *p = new int[matrix_num];
    cout << "请输入各矩阵的列数：" << endl;
    for (int i = 0; i < matrix_num; i++)
    {
        cin >> p[i];
    }

    matrix_chain(p, matrix_num, m, s);
    traceback(0, matrix_num - 1, s);

    system("pause");
    return 1;
}

可结合我的另一篇关于贪心算法的博客进行比较，了解这两者的区别；

（http://www.cnblogs.com/zf-blog/p/8674932.html）

原文地址：https://www.cnblogs.com/zf-blog/p/8763090.html