zoukankan      html  css  js  c++  java
  • hdoj--2119--Matrix(最小点覆盖)

    Matrix

    Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 2319    Accepted Submission(s): 1030


    Problem Description
    Give you a matrix(only contains 0 or 1),every time you can select a row or a column and delete all the '1' in this row or this column .

    Your task is to give out the minimum times of deleting all the '1' in the matrix.
     

    Input
    There are several test cases.

    The first line contains two integers n,m(1<=n,m<=100), n is the number of rows of the given matrix and m is the number of columns of the given matrix.
    The next n lines describe the matrix:each line contains m integer, which may be either ‘1’ or ‘0’.

    n=0 indicate the end of input.
     

    Output
    For each of the test cases, in the order given in the input, print one line containing the minimum times of deleting all the '1' in the matrix.
     

    Sample Input
    3 3 0 0 0 1 0 1 0 1 0 0
     

    Sample Output
    2
     

    Author
    Wendell
     

    Source

    给了一个n*m的矩阵,矩阵元素有0 1,每次可以消除一行或者一列的1,问最少多少次消除全部的1
    最少点覆盖=最大匹配,匈牙利算法:
    #include<stdio.h>
    #include<vector>
    #include<string.h>
    #include<algorithm>
    using namespace std;
    vector<int>map[200];
    int pipei[200],used[200];
    int find(int x)
    {
    	for(int i=0;i<map[x].size();i++)
    	{
    		int y=map[x][i];
    		if(!used[y])
    		{
    			used[y]=1;
    			if(!pipei[y]||find(pipei[y]))
    			{
    				pipei[y]=x;
    				return 1;
    			}
    		}
    	}
    	return 0;
    }
    int main()
    {
    	int n,m;
    	while(scanf("%d",&n),n)
    	{
    		scanf("%d",&m);
    		for(int i=1;i<=n;i++)
    		{
    			map[i].clear();
    			pipei[i]=0;
    		}
    		int a;
    		for(int i=1;i<=n;i++)
    		{
    			for(int j=1;j<=m;j++)
    			{
    				scanf("%d",&a);
    				if(a)
    					map[i].push_back(j);
    			}
    		}
    		int sum=0;
    		for(int i=1;i<=n;i++)
    		{
    			memset(used,0,sizeof(used));
    			sum+=find(i);
    		}
    		printf("%d
    ",sum);
    	}
    }


  • 相关阅读:

    暴力求解/数学问题
    Leetcode207. Course Schedule
    Balanced Team
    由先序和中序求后序
    Median String
    树的同构
    uva 202
    整除光棍
    阅览室
  • 原文地址:https://www.cnblogs.com/playboy307/p/5273491.html
Copyright © 2011-2022 走看看