To the Max

[daolaAmeng]

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

[bargio_susie]

搜索最大子段和。

[f22fbi]

E文不太好
要求和最大的矩阵
这个矩阵是任意位置的吗？
不一定是例子那样，从左下角开始？

[c_spark]

#include <stdio.h>

#define MAX 130

int maxSum(int b[], int n)
{
	int max = b[0];
	int sum = 0;
	for(int i = 0; i < n; ++i)
	{
		if(sum < 0) sum = b[i];
		else sum += b[i];
		if(sum > max) max = sum;
	}
	return max;
}

int solve(int a[][MAX], int n)
{
	int b[MAX];
	int i,j,k,sum,tmp;
	sum = a[0][0];
	for(i = 0; i < n; ++i)
	{
		for(k = 0; k < n; ++k) b[k] = 0;
		for(j = i; j < n; ++j)
		{
			for(k =0; k < n; ++k) b[k] += a[j][k];
			if(sum < (tmp = maxSum(b,n)) ) sum = tmp;
		}
	}
	return sum;
}

int main()
{
	int arr[MAX][MAX];
	int i,j,n;
	while(scanf("%d",&n)!=EOF)
	{
		for(i = 0; i < n; ++i)
			for(j = 0; j < n; ++j)
				scanf("%d",&arr[i][j]);
		printf("%d\n",solve(arr,n));
	}
	return 0;
}

详细过程我作了分析
仅供参考...
http://blog.csdn.net/c_spark/archive/2008/04/26/2330336.aspx