• 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

    15

    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;
    }



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  • 原文地址:https://www.cnblogs.com/lightspeedsmallson/p/4785843.html
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