zoukankan      html  css  js  c++  java
  • Hdoj 1213.How Many Tables 题解

    Problem Description

    Today is Ignatius' birthday. He invites a lot of friends. Now it's dinner time. Ignatius wants to know how many tables he needs at least. You have to notice that not all the friends know each other, and all the friends do not want to stay with strangers.

    One important rule for this problem is that if I tell you A knows B, and B knows C, that means A, B, C know each other, so they can stay in one table.

    For example: If I tell you A knows B, B knows C, and D knows E, so A, B, C can stay in one table, and D, E have to stay in the other one. So Ignatius needs 2 tables at least.

    Input

    The input starts with an integer T(1<=T<=25) which indicate the number of test cases. Then T test cases follow. Each test case starts with two integers N and M(1<=N,M<=1000). N indicates the number of friends, the friends are marked from 1 to N. Then M lines follow. Each line consists of two integers A and B(A!=B), that means friend A and friend B know each other. There will be a blank line between two cases.

    Output

    For each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks.

    Sample Input

    2
    5 3
    1 2
    2 3
    4 5
    
    5 1
    2 5
    

    Sample Output

    2
    4
    

    Author

    Ignatius.L

    Source

    杭电ACM省赛集训队选拔赛之热身赛


    思路

    裸并查集题目,上模板就好了QAQ

    代码

    #include<bits/stdc++.h>
    using namespace std;
    int father[110];
    void init(int n)
    {
        for(int i=1;i<=n;i++) father[i]=i;
    }
    int find(int x)
    {
        while(father[x]!=x) x=father[x];
        return x;
    }
    void join(int a,int b)
    {
        int t1=find(a);
        int t2=find(b);
        if(t1!=t2) father[t1]=t2;
    }
    int getNum(int n)
    {
    	int num = 0;
    	for(int i=1;i<=n;i++)
    		if(father[i]==i)
    			num++;
    	return num;
    }
    int main()
    {
    	int t;
    	cin >> t;
    	while(t--)
    	{
    		int n,m;
    		cin >> n >> m;
    		init(n);
    		for(int i=1;i<=m;i++)
    		{
    			int a,b;
    			cin >> a >> b;
    			if(find(a) != find(b))
    				join(a,b);
    		}
    		cout << getNum(n) << endl;
    	}
    	return 0;
    }
    
  • 相关阅读:
    Count on a tree
    图论1 1009
    DP2 1008
    DP1 1008
    NOIP 模拟 1006
    2019 CSP-S 初赛退役记
    9.13——TEST NOIP模拟测试
    [洛谷P2387][NOI2014]魔法森林
    [洛谷P2596][ZJOI2006]书架
    [BZOJ4241]历史研究
  • 原文地址:https://www.cnblogs.com/MartinLwx/p/10042271.html
Copyright © 2011-2022 走看看