zoukankan      html  css  js  c++  java
  • hdu1213 How Many Tables

    How Many Tables

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

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 9188    Accepted Submission(s): 4510


    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
    简单的并查集
    #include<iostream>
    #include<string.h>
    #include<stdio.h>
    using namespace std;
    int father[1010],num[1010];
    int Find(int x)
    {
        int r=x,tmp;
        while(r!=father[r])
        {
            r=father[r];
        }
        while(x!=r)
        {
            tmp=father[x];
            father[x]=r;
            x=tmp;
        }
        return r;
    }
    void Union(int a,int b)
    {
        a=Find(a);b=Find(b);
        if(a!=b)
        {
            father[a]=b;
            num[b]+=num[a];
            num[a]=0;
        }
    }
    int main()
    {
        int t,n,m,i,j;
        cin>>t;
        while(t--)
        {
            cin>>n>>m;
            for(i=1;i<=n;i++)
            {
                 father[i]=i;num[i]=1;
            }
    
            int p1,p2;
            for(i=1;i<=m;i++)
            {
                scanf("%d%d",&p1,&p2);
                Union(p1,p2);
            }
             int sum=0,sum2=0;
             for(i=1;i<=n;i++)
               if(num[i]!=0)
                  {
                      sum++;
                      sum2+=num[i];
                  }
    
             cout<<n-sum2+sum<<endl;
        }
    }
    View Code
  • 相关阅读:
    判断元素的属性是否存在
    js 查找树节点 数组去重
    redis 基础知识
    jQuey知识点三 解析json数据
    jQuery知识点二 实现隔行变色
    mysql 基础操作一
    ruby 基础知识三 读写文件
    Active Record 数据迁移
    ruby 基础知识(二)
    rails 常用的知识点
  • 原文地址:https://www.cnblogs.com/ainixu1314/p/3217560.html
Copyright © 2011-2022 走看看