Problem Description

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.

 

Input

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.

 

Sample Input

4

0 -2 -7 0 9 2 -6 2

-4 1 -4 1 -1

8 0 -2

 

Sample Output

15

 

题目大意：给一个N*N的矩阵求解最大的子矩阵和

解法：压缩数组+暴力（水过）

源代码：

<span style="font-size:18px;">#include<iostream>

#include<stdio.h>

#include<stdlib.h>

#include<string>

#include<string.h>

#include<math.h>

#include<map>

#include<vector>

#include<algorithm>

#include<queue>

using namespace std;

#define MAX 0x3f3f3f3f

#define MIN -0x3f3f3f3f

#define PI 3.14159265358979323

#define N 105

int n;

int ans[N][N];

int value(int x, int y)

{

	int sum;

	int i, j;

	sum = 0;

	for (i = 1; i <= n; i++)

	{

		for (j = 1; j <= n; j++)

		{

			if (i >= x&&j >= y)

				sum = max(sum, ans[i][j] + ans[i - x][j - y] - ans[i - x][j] - ans[i][j - y]);

			if (i >= y&&j >= x)

				sum = max(sum, ans[i][j] + ans[i - y][j - x] - ans[i - y][j] - ans[i][j - x]);

		}

	}

	return sum;

}

int main()

{

	int i, j;

	int result;

	int num;

	int temp;

	while (scanf("%d", &n) != EOF)

	{

		memset(ans, 0, sizeof(ans));

		for (i = 1; i <= n; i++)

		{

			for (j = 1; j <= n; j++)

			{

				scanf("%d", &num);

				ans[i][j] = ans[i - 1][j] + ans[i][j - 1] - ans[i - 1][j - 1] + num;

			}

		}

		result = 0;

		for (i = 1; i <= n; i++)

		{

			for (j = i; j <= n; j++)

			{

				temp = value(i, j);

				if (temp > result)

					result = temp;

			}

		}

		printf("%d\n", result);

	}

	return 0;

}</span>