How Many Tables（并查集模板）

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

题目大概意思；

今天是伊格纳修斯的生日。他邀请了很多朋友。现在是吃饭时间了。伊格纳修斯想知道他至少需要多少张桌子。你必须注意到，不是所有的朋友都认识对方，所有的朋友都不想和陌生人呆在一起。

这个问题的一个重要规则是，如果我告诉你A知道B，B知道C，这意味着A，B，C彼此认识，所以他们可以留在一张桌子里。

例如：如果我告诉你A知道B，B知道C，D知道E，所以A，B，C可以留在一张桌子里，而D，E必须留在另一张桌子上。所以伊格纳修斯至少需要两张桌子。

输入

输入以表示测试案例数量的整数 T（1<=T <=25）开头。然后T测试案例如下。每个测试案例以两个整数 N 和 M（1<=N，M<=1000）开头。N 表示好友数量，好友标记为 1 到 N。然后M线跟随。每行由两个整数 A 和 B （A！=B） 组成，这意味着朋友 A 和朋友 B 彼此认识。两种情况之间将有一条空白线。

输出

对于每个测试案例，只需输出 至少需要多少桌子。不要打印任何空白。

分析；就是一个模板题，找有几个根的事情；

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上ac代码。

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学习算法

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+6;
int f[maxn];
int find(int x)//根
{
  if(f[x]!=x)return find(f[x]);return f[x];
}
int main()
{
  int t;cin>>t;
  while(t--)
  {
    int n,m,a,b,ans=0;
    cin>>n>>m;
    for(int i=1;i<=n;i++)
    f[i]=i;
    for(int i=1;i<=m;i++)//合并
    {
      cin>>a>>b; 
      int r1=find(a);
      int r2=find(b);
      if(r1!=r2)
      f[r1]=r2;
    }
    for(int i=1;i<=n;i++)
    {
    if(f[i]==i)//有几个根 
    {
      ans++;
    }
    }
    cout<<ans<<endl;  
  }
}