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

#include<stdio.h>

#include<cmath>

#include<iostream>

#define INF 0x3f3f3f3f

#define me(a,b) memset(a,b,sizeof(a))

#define N 102

typedef long long ll;

using namespace std;

int n,a[N][N],dp[N][N],t,sum,ans;

int main()

{

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

      cin>>n;

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

      {

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

            {

                  cin>>t;

                  a[i][j]=a[i][j-]+t;//记录前缀和

            }

      }

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

      {

            for(int j=i;j<=n;j++) //枚举，第i列道第j列

            {

                  sum=;

                  for(int k=;k<=n;k++)//最大子序列遍历一遍

                  {

                        int t=a[k][j]-a[k][i-];//t代表： 第k行中，第 i 列到第 j 列的和

                        sum+=t;

                        sum=sum<?:sum; //<0则置为0

                        if(sum>ans)

                              ans=sum;

                  }

            }

      }

      cout<<ans;

}