POJ 1050：To the Max

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 43241		Accepted: 22934

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

题意是给定一个矩阵，求其子矩阵的最大和。

这题也是弄得相当郁闷，一开始暴力，结果预料之中的TLE。然后试了一下dp，结果还MLE。。。郁闷得不行。

然后看了别人的思路，发现可以二维变一维，想了想忽然恍然大悟。

将每一列的加起来，就是一维了。枚举不同行即可。之前怎么做的这次怎么求。

代码：

#include <iostream>

#include <string>

#include <cstring>

#include <algorithm>

#include <cmath>

#pragma warning(disable:4996)

using namespace std;

int value[250][250];

int value2[250];

int dp[250];

int main()

{

	//freopen("input.txt","r",stdin);

	//freopen("out.txt","w",stdout);

	int N,i,j,h,k,g,f;

	int ans=-100;

	scanf("%d",&N);

	memset(dp,0,sizeof(dp));

	memset(value2,0,sizeof(value2));

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

	{

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

		{

			scanf("%d",&value[i][j]);

			ans=max(ans,value[i][j]);

		}

	}

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

	{

		for(h=i;h<=N;h++)

		{

			for(k=1;k<=N;k++)

			{

				value2[k] += value[h][k];

				dp[k] = max(dp[k-1]+value2[k],value2[k]);

				ans = max(ans,dp[k]);

			}

			memset(dp,0,sizeof(dp));

		}

		memset(value2,0,sizeof(value2));

	}

	cout<<ans<<endl;

	return 0;

}

巴特西

POJ 1050：To the Max

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