poj1050 To the Max 最大子矩阵

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 -2Sample Output

15

最大子矩阵，首先一行数列很简单求最大的子和，我们要把矩阵转化成一行数列，就是从上向下在输入的时候取和，map[i][j]表示在J列从上向下的数和，这样就把一列转化成了一个点，再用双重，循环，任意i行j列开始的一排数的最大和，就是最终的最大和

#include  
#include  
using namespace std; 
int map[105][105]; 
int main() 
{ 
    int n,i,j,k,sum,x,max; 
    while(scanf("%d",&n)!=EOF) 
    { 
        for(i=0;i
#include 
using namespace std;
int map[105][105];
int main()
{
    int n,i,j,k,sum,x,max;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=0;i 
 
             sum=0;
                     if(sum>max)
                         max=sum;
 
 
                 }
             }
             printf("%d\n",max);
     }
 
     return 0;
 }
 
                         sum=0;
                     if(sum>max)
                         max=sum;
 
 
                 }
             }
             printf("%d\n",max);
     }
 
     return 0;
 }