动态规划之To the Max

题目：

Description

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

这道题一开始想歪了....后来看了别人的，发现只是把一个二维的压缩成一维的，然后问题就转化为求最大子段和的问题了

代码如下（代码写的好捉急的说...）：

#include
int num[100][100];
int sum[100];
int f[100];
int N;
int sum_col(int i,int j,int col){
    int ans,row;
    ans=0;
    for(row=i;rowtmp)
                    tmp=f[k];
            }
            if(tmp>ans)
                ans=tmp;
        }
    printf("%d\n",ans);
    return 0;
}