Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle. 
As an example, the maximal sub-rectangle of the array:

0 -2 -7 0 
9 2 -6 2 
-4 1 -4 1 
-1 8 0 -2 
is in the lower left corner:

9 2 
-4 1 
-1 8 
and has a sum of 15. 

The input consists of an N * N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N^2 integers separated by whitespace (spaces and newlines). These are the N^2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will be in the range [-127,127].

Output the sum of the maximal sub-rectangle.

4

0 -2 -7 0 9 2 -6 2

-4 1 -4  1 -1

8  0 -2

15

和最大连续子串和类似，可以把二维的转化为一维，将i行到j行的各列数分别相加，求最大连续子串和即可。
同时可利用前缀和，sum[i][j]表示第j列前i行相加的和，i从1开始，则i-j行相加的和为sum[j][k]-sum[i-1][k]

#include <iostream>

#include <stdio.h>

#include <cstring>

#include <algorithm>

using namespace std;

#define INF 0x3f3f3f3f

int a[][];

int t[], dp[];

int sum[][];

int n;

int sovle()

{

    int ans = -INF;

    for (int i = ; i <= n; i++) {

        for (int j = i; j <= n; j++) {

            int Max;

            dp[] = sum[j][] - sum[i-][];

            for (int k = ; k <= n; k++) {

                int temp = sum[j][k]-sum[i-][k];

                dp[k] = max(dp[k-]+temp, temp);

            }

            Max = *max_element(dp+, dp+n+);

            if (Max > ans)

                ans = Max;

        }

    }

    return ans;

}

int main()

{

    //freopen("1.txt", "r", stdin);

    scanf("%d", &n);

    memset(sum, , sizeof(sum));

    for (int i = ; i <= n; i++)

        for (int j = ; j <= n; j++) {

            scanf("%d", &a[i][j]);

            sum[i][j] = sum[i-][j] + a[i][j];

        }

    printf("%d\n", sovle());

    return ;

}