poj1050

i, j 代表列

用前缀和预处理出i,j之间的和

枚举i,j　那么i到j之间的和就相当于最长连续字段和的一位，对n行都做相同处理，再做一遍最长连续字段和即可。

最长连续字段和的 dp方程　dp[i] = max(dp[i-1] + a[i], a[i]);

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int n, a[110][110], sum[110][110], tmp[110], dp[110], ans;
int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            scanf("%d", &a[i][j]);
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            sum[i][j] += sum[i][j-1] + a[i][j];
    for (int i = 1; i <= n; i++)
        for (int j = i; j <= n; j++){
            memset(dp, 0, sizeof(dp));
            for (int k = 1; k <= n; k++)
            {
                tmp[k] = sum[k][j] - sum[k][i-1];
            }
            for(int l = 1; l <= n; l++)
            {
                dp[l] = max(dp[l-1] + tmp[l], tmp[l]);
                if(dp[l] > ans)ans = dp[l];
            }
        }
    printf("%d
", ans);
}