DP--ZOJ - 1074 To the Max

Description

Problem

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 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.

The input consists of an N x 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.

Example

Input

4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2

Output

15 

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 110;
const int INF = 0x3f3f3f3f;
int a[maxn][maxn], d[maxn], n;

int dp(int* d)
{
	int sum = 0, mx = -INF;
	for(int i=0; i<n; i++) {
        sum += d[i];
        if(sum < 0) sum = d[i];
        mx = max(mx, sum);
	}
	return mx;
}

int main()
{
	while(~scanf("%d", &n)) {
		for(int i=0; i<n; i++)
			for(int j=0; j<n; j++)
				scanf("%d", &a[i][j]);
		int sum = -INF;
		for(int i=0; i<n; i++) {
			memset(d, 0, sizeof(d));
			for(int j=i; j<n; j++) {
				for(int k=0; k<n; k++)
					d[k] += a[j][k];
				sum = max(sum, dp(d));
			}
		}
		printf("%d\n", sum);
	}
    return 0;
}