动态规划——F 最大矩阵和

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

解题思路：
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，a[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
注意：如果m<0，则就不需要继续加了
程序代码：

#include <cstdio>
#include <cstring>
using namespace std;
int a[110][110];
int main()
{
    int n,c;
    while( scanf("%d",&n)==1)
    {
      memset(a,0,sizeof(a));
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            {
                scanf("%d",&c);
                a[i][j]=a[i-1][j]+c;
            }
    int sum=0;
    for(int i=1;i<=n;i++)
        for(int j=i;j<=n;j++)
        {
            int m=0;
            for(int k=1;k<=n;k++)
            {
                int t=a[j][k]-a[i-1][k];
                m+=t;
                if(m<0) m=0;
                if(sum<m) sum=m;
            }
        }

    printf("%d
",sum);
    }
    return 0;
}