动态规划之最大子矩阵

步骤

1. 枚举两行（i , j）

2.将两行压缩，转换为最大子段和

小技巧

#include <iostream>
#include <memory.h>
int n;
using namespace std;
int a[105][105];
int sum[105][105];
int dp[105][105][105];
int main()
{
    cin >> n;
    for( int i = 1; i <= n; i++ )
        for( int j = 1; j <= n; j++ )
        {
            cin >> a[i][j];
            for( int k = 1; k <= n; k++)
                dp[i][j][k] = -(1<<30);
        }
    memset(sum, 0, sizeof(a));
    for( int i = 1; i <= n; i++ )
        for( int j = 1; j <= n; j++ )
        {
            sum[i][j] = sum[i-1][j] + a[i][j];
        }
    for( int i = 1; i <= n; i++)
        for( int j = i; j <= n; j++)
        {
            for(int k = 1; k <= n; k++)
                dp[i][j][k] = max(dp[i][j][k-1] + sum[j][k] - sum[i][k], sum[j][k] - sum[i][k] );
        }
    int Max = -(1<<30);
    for( int i = 1; i <= n; i++)
        for( int j = i; j <= n; j++)
            for(int k = 1; k <= n; k++)
                Max = max(Max, dp[i][j][k]);
    cout << Max << endl;
    return 0;
}