poj_1050To the Max

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 36015		Accepted: 18897

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

果将每一个二维矩阵的每一列的数据总和,按照列的序号分别对应地保存到一个一维数组中,就转化成一维,其性质跟求最大子段和一样,而在这里只不过要进行多次这样的计算而已 ^ -^ 经典Dynamic programming 中的经典!!!!

#include <iostream>
#include <cstring>
using namespace std;
#pragma warning(disable : 4996)
const int MAXN = 105;
int arr[MAXN][MAXN], dp[MAXN];
int n;

int DP()
{
	int thissum, maxsum;
	thissum = maxsum = 0;
	for (int i = 1; i <= n; i++)
	{
		thissum += dp[i];
		if(thissum > maxsum)
		{
			maxsum = thissum;
		}
		if(thissum < 0)
		{
			thissum = 0;
		}
	}
	return maxsum;
}

int main()
{
	freopen("in.txt", "r", stdin);
	int i, j, k;
	int sum, ans;
	while(scanf("%d", &n) != EOF)
	{
		for(i = 1; i <= n; i++)
		{
			for(j = 1; j <= n; j++)
			{
				scanf("%d", &arr[i][j]);
			}
		}
		ans = 0;
		for (int i = 1; i <= n; i++)
		{
			memset(dp, 0, sizeof(dp));
			for (int j = i; j <= n; j++)
			{
				for (k = 1; k <= n; k++)
				{
					dp[k] += arr[j][k];
				}
				sum = DP();
				if(sum > ans)
				{
					ans = sum;
				}
			}
		}
		printf("%d\n", ans);
	}
}