zoukankan      html  css  js  c++  java
  • [kuangbin带你飞]专题五 并查集 C

    C - How Many Tables

    题目链接:https://vjudge.net/contest/66964#problem/C

    题目:

    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.

    InputThe 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.
    OutputFor 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
    题意:说有n个人要来参加生日聚会,认识的人坐在一个桌上,如果A->B,B->C,那么ABC坐在一个桌上,求有多少个桌子
    思路:找每个人的“爸爸”即可,看有多少个“爸爸”就是有多少个桌子辽,用set存储“爸爸”即可,输出st.size(),即就是答案,注意下标从1开始,一开始俺从0开始俺就WA了一发
    //
    // Created by hy on 2019/7/23.
    //
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <queue>
    #include<set>using namespace std;
    const int maxn=3e4+10;
    int father[maxn];
    int find(int x)
    {
        if(x==father[x])
            return x;
        return father[x]=find(father[x]);
    }
    void merge(int x,int y)
    {
        int xx=find(x);
        int yy=find(y);
        if(xx!=yy)
            father[xx]=yy;
    }
    int main()
    {
        int T;
        scanf("%d",&T);
        while(T--)
        {
            int n,m;
            scanf("%d%d",&n,&m);
            for(int i=1;i<=n;i++)
                father[i]=i;
            int x,y;
            for(int i=0;i<m;i++)
            {
                scanf("%d%d",&x,&y);
                merge(x,y);
            }
            set<int>st;
            for(int i=1;i<=n;i++)
            {
                int cun=find(i);
                st.insert(cun);
            }
            printf("%d
    ",st.size());
        }
        return 0;
    }
    
    
  • 相关阅读:
    Python股票分析系列——系列介绍和获取股票数据.p1
    快速相关
    特别长序列的快速卷积
    长序列的快速卷积
    快速卷积
    素因子快速傅里叶变换
    用一个N点复序列的FFT同时计算两个N点实序列离散傅里叶变换
    实序列快速傅里叶变换(二)
    实序列快速傅里叶变换(一)
    java 对于手机号码、邮箱、银行卡号脱敏一条龙服务
  • 原文地址:https://www.cnblogs.com/Vampire6/p/11234208.html
Copyright © 2011-2022 走看看