To the max(求最大子矩阵和)

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 47985		Accepted: 25387

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

枚举：单纯枚举必然超时，n^6完全不可取

前缀和优化枚举：分别搜一个左上角的点和一个右上角的点，可以将复杂度降到n^4，但仍然超时

于是我们考虑，可不可以换一种思路：

可以枚举上下边界，然后问题就转变成了一位字段和求最大

#include<iostream>
using namespace std;
int n,map[200][200],sum[200][200],ans;
int main(){
    cin>>n;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++){
            cin>>map[i][j];
            sum[i][j]=sum[i][j-1]+map[i][j];
        }
    for(int i=1;i<n;i++){
        for(int j=i;j<=n;j++){
            int Sum=0;
            for(int k=1;k<=n;k++){
                Sum+=sum[k][j]-sum[k][i-1];
                if(Sum<0)Sum=0;
                else if(Sum>ans)ans=Sum;
            }
        }
    }
    cout<<ans;
}