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[][];
 int main()
 {
     int n,c;
     while( scanf("%d",&n)==)
     {
       memset(a,,sizeof(a));
     for(int i=;i<=n;i++)
         for(int j=;j<=n;j++)
             {
                 scanf("%d",&c);
                 a[i][j]=a[i-][j]+c;
             }
     int sum=;
     for(int i=;i<=n;i++)
         for(int j=i;j<=n;j++)
         {
             int m=;
             for(int k=;k<=n;k++)
             {
                 int t=a[j][k]-a[i-][k];
                 m+=t;
                 if(m<) m=;
                 if(sum<m) sum=m;
             }
         }

     printf("%d\n",sum);
     }
     return ;
 }