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題目鏈接：

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

How Many Tables

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 41546 Accepted Submission(s): 20798

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 杭電ACM省賽集訓隊選拔賽之熱身賽 代碼如下：

#include<stdio.h>
#include<iostream>
using namespace
 
 std;
#define max_v 50005
int pa[max_v];//pa[x] 表示x的父節點
int rk[max_v];//rk[x] 表示以x為根結點的樹的高度
int n,ans;
void make_set(int x)
{
    pa[x]=x;
    rk[x]=0;//一開始每個節點的父節點都是自己
}
int find_set(int x)//帶路徑壓縮的查找
{
    if(x!=pa[x])
        pa[x]=find_set(pa[x]);
    return pa[x];
}
void union_set(int x,int y)
{
    x=find_set(x);//找到x的根結點
    y=find_set(y);
    if(x==y)//根結點相同 同一棵樹
        return ;
    ans--;
    if(rk[x]>rk[y])
    {
        pa[y]=x;
    }
    else
    {
        pa[x]=y;
        if(rk[x]==rk[y])
            rk[y]++;
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,m;
        scanf("%d %d",&n,&m);
        if(m+n==0)
            break;
        for(int i=1; i<=n; i++)
        {
            make_set(i);
        }
        ans=n;
        for(int i=0; i<m; i++)
        {
            int x,y;
            scanf("%d %d",&x,&y);
            union_set(x,y);
        }
        printf("%d\n",ans);
    }
    return 0;
}

HDU 1213（裸並查集）（無變形）