POJ 1050 - To the Max

题目链接：http://poj.org/problem?id=1050

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

求最大子矩阵元素和，从数据量上看，可以发现非常小，不难猜想是一道比较暴力的题目。

听说是DP？我也不知道求矩阵前缀和算不算DP……

我们用dp[i][j]表示这个矩阵前i行，前j列所有元素的和。

那么，我们求任何一个子矩阵 [x1,y1][x2,y2] ，就用dp[x2][y2]-dp[x2][y1]-dp[x1][y2]+dp[x1 - 1][y1 - 1]表示即可；

然后，直接四重循环枚举x1,x2,y1,y2就可得到答案。

#include<cstdio>
#include<cstring> 
#define MAXN 105
#define INF 0x3f3f3f3f
int n,dp[MAXN][MAXN];
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                int tmp;
                scanf("%d",&tmp);
                dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]+tmp;
            }
        }
        int ans=-INF;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                for(int ii=i;ii<=n;ii++)
                {
                    for(int jj=j;jj<=n;jj++)
                    {
                        int now=dp[ii][jj]-dp[ii][j-1]-dp[i-1][jj]+dp[i-1][j-1];
                        if(ans<now) ans=now;
                    }
                }
            }
        }
        printf("%d\n",ans);
    }
}