zoukankan      html  css  js  c++  java
  • hdu-1213-How Many Tables

    How Many Tables

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 25116    Accepted Submission(s): 12523


    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
     

    题目大意:水题计算最后的集合个数。

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1213

    代码:

    #include <iostream>
    
    using namespace std;
    int f[1005];
    int find(int n)
    {
        if(n!=f[n])
        f[n]=find(f[n]);
        return f[n];
    }
    int main()
    {   int t,n,m;
        cin>>t;
    while(t--)
    {
        cin>>n>>m;
        for(int i=1;i<=n;i++)
        f[i]=i;
        int a,b;
        for(int i=0;i<m;i++)
        {
            cin>>a>>b;
            int a1=find(a);
            int b1=find(b);
            f[a1]=b1;
        }
        int t1=0;
        for(int i=1;i<=n;i++)
        if(find(i)==i)
        t1++;
        cout<<t1<<endl;
    }
    
    }
  • 相关阅读:
    前端程序员容易忽视的一些基础知识
    一道前端学习题
    Unity调用Windows对话框保存时另存为弹框
    Unity镜子效果的实现(无需镜子Shader)
    Unity射线检测的用法总结
    unity中实现简单对象池,附教程原理
    Unity调用Window提示框Yes/No(英文提示窗)
    Unity调用Windows弹框、提示框(确认与否,中文)
    C#LinQ语法
    服务器的购买与网站的创建
  • 原文地址:https://www.cnblogs.com/llsq/p/5880986.html
Copyright © 2011-2022 走看看