F - To the Max

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.

Input

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

Output the sum of the maximal sub-rectangle.

Sample Input

4
0 -2 -7 0 9 2 -6 2
-4 1 -4  1 -1

8  0 -2

Sample Output

15

#include<cstdio>
#include<string.h>
using namespace std;
int map[120][120];
int main()
{
    int n;
    int flag;
    int sum;
    while(scanf("%d",&n)!=EOF)
    {
        memset(map,0,sizeof(map));
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
            {
                scanf("%d",&flag);
                map[i][j]+=map[i][j-1]+flag;//先计算每一行的前J列的和
        for(int i=1; i<=n; i++)
            for(int j=1; j<=i; j++)//确定子矩阵的左右边界
            {
                int sum=0;
                for(int k=1; k<=n; k++)//行数
                {
                    if(sum<0)
                        sum=0;//如果前几行的和是小于0的，就令sum=0;因为负数越加越小。
                    sum+=(map[k][i]-map[k][j-1]);//一行一行地加起来，第k行的[i,j]列
                    if(sum>max)
                        max=sum;//找出最大值！！！
                }
                            
            }
            printf("%d\n",max);
    }
    return 0;
}