杭电oj--1213 How Many Tables

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

题目大意：第一行输入t，表示有t行测试实例，在输入n和m，表示有n各朋友，m对朋友。

  Ignatiu的朋友有一个原则：  他们是不会和陌生人在一个桌子上的。即 朋友的朋友是朋友，那么可以坐一个桌子上，问至少要有多少个桌子？

解题思路：并查集判断图连通，或者找出图有多少个连通分量

#include 
using namespace std;
#define N 1001
int father[N];
void init()//初始化并查集
{
	for(int i=0;i>t;
	while(t--)
	{
		cin>>n>>m;
		init();
		while(m--)
		{
			cin>>a>>b;
			int cx=getfather(a);
			int cy=getfather(b);
			if(cx!=cy) {father[cx]=cy;}//合并两个元素所在的集合
		}
		int count=0;
		for(i=1;i<=n;i++)
		{
			if(father[i]==i)//扫描一遍father数组
			{	count++;}
		}
		cout<