hdu 1213 How Many Tables

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

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

#include <iostream>
#include <algorithm>
#include <cmath>
#include <stdio.h>
#include <cstring>
#include <string>
#include <cstdlib>
#include <queue>
#include <stack>
#include <set>
#include <vector>
#include <map>
#include <list>
#include <iomanip>
 #include <fstream>

using namespace std;
int fa[1009];
//这题一看就感觉和寒假的poj 2524异曲同工。都是普通的并查集 
int get(int x)//找朋友总代表 
{
    int root=x;
    while(fa[root]!=root)//不断向上寻找根节点 
        root=fa[root];
     
    int t=x,temp;
    while(fa[t]!=root)//自底（x)向上（root)更新一波 
    {
        temp=fa[t];//感觉swap函数并不可靠，之前用了换成这样朴素的交换就从WA变成AC了 
        fa[t]=root;
        t=temp;
    }
     return root;     
}

void join(int x,int y)//保存朋友关系 
{
    int gx=get(x),gy=get(y);
    if(gx!=gy)//x 和 y的根节点不同 
        fa[gx]=gy;//那么这两者都归入到x根节点的根节点（大概就是最顶端的那个） 
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m,x,y,cnt=0;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;++i)
        {
            fa[i]=i;
        }
        for(int i=1;i<=m;++i)
        {
            scanf("%d%d",&x,&y);
            join(x,y);
        }
        for(int i=1;i<=n;++i)
        {
            if(get(i)==i)
                cnt++;
        }
        printf("%d
",cnt);
    }
    return 0;
}