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[][];
 int main()
 {
     int n;
     int flag;
     int sum;
     while(scanf("%d",&n)!=EOF)
     {
         memset(map,,sizeof(map));
         for(int i=; i<=n; i++)
             for(int j=; j<=n; j++)
             {
                 scanf("%d",&flag);
                 map[i][j]+=map[i][j-]+flag;//先计算每一行的前J列的和
         for(int i=; i<=n; i++)
             for(int j=; j<=i; j++)//确定子矩阵的左右边界
             {
                 int sum=;
                 for(int k=; k<=n; k++)//行数
                 {
                     if(sum<)
                         sum=;//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-]);//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 ;
 }