zoukankan      html  css  js  c++  java
  • [ACM] hdu 1213 How Many Tables(并查集)

    How Many Tables

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 11580    Accepted Submission(s): 5696


    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


    解题思路:

    简单的并查集。计算有多少集合。

    代码:

    #include <iostream>
    using namespace std;
    int parent[1002];
    
    void init(int n)
    {
        for(int i=1;i<=n;i++)
            parent[i]=i;
    }
    int find(int x)
    {
        return parent[x]==x?x:find(parent[x]);
    }
    void unite(int x,int y)
    {
        x=find(x);
        y=find(y);
        if(x==y)
            return ;
        else
            parent[x]=y;
    }
    int main()
    {
        int t;cin>>t;
        int n,m;
        int a,b;
        while(t--)
        {
            cin>>n>>m;
            init(n);
            int ans=0;
            for(int i=1;i<=m;i++)
            {
                cin>>a>>b;
                unite(a,b);
            }
            for(int i=1;i<=n;i++)
                if(parent[i]==i)
                ans++;
            cout<<ans<<endl;
        }
        return 0;
    }
    


  • 相关阅读:
    IT题库-134 | String、StringBuffer和StringBuilder的区别
    Java NIO 总结
    Java 性能分析工具-JProfiler
    Java 性能分析工具-MAT
    java 获取dump文件
    GC总结
    String总结
    堆-对象的分配与回收过程
    java 生产环境调优排查总结
    记一次因为Gradle与Lombok不兼容导致编译时的内存溢出 Expiring Daemon because JVM heap space is exhausted
  • 原文地址:https://www.cnblogs.com/sr1993/p/3697913.html
Copyright © 2011-2022 走看看