POJ ---1050 To the Max

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 38914		Accepted: 20534

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

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int map[105][105], sum[105][105];
int main(){
    int n, maxn;
    /* freopen("in.c", "r", stdin); */
    while(~scanf("%d", &n)){
        for(int i = 1;i <=n;i ++){
            for(int j = 1;j <= n;j ++){
                scanf("%d", &map[i][j]);
                sum[i][j] = sum[i-1][j] + sum[i][j-1] + map[i][j] - sum[i-1][j-1];
            }
        }
        maxn = -100000000;
        for(int i = 1;i <= n;i ++){
            for(int j = 1;j <= n;j ++){
                for(int k = i+1;k <= n;k ++){
                    for(int p = j+1;p <= n;p ++)
                        maxn = max(sum[k][p]-sum[k][j-1]-sum[i-1][p] + sum[i-1][j-1], maxn);
                }
            }
        }
        printf("%d\n", maxn);
    }
    return 0;
}