How Many Tables （并查集）

     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

题意：一个人办聚会，要来的人不是朋友的不能坐一桌，坐一桌的朋友可以是间接的朋友。问最少需要准备多少张桌子。

分析：利用并查集找出共有多少个祖先，即需要准备的桌子数量。

#include<stdio.h>
int f[1001],m,n;
void init()
{
	int i;
	for(i = 1; i <= n; i ++)
		f[i] = i;
	return ;
}
int getf(int v)
{
	if(f[v] == v)
		return v;
	else
	{
		f[v] = getf(f[v]);
		return f[v];
	}
}
void merge(int v, int u)
{
	int t1,t2;
	t1 = getf(v);
	t2 = getf(u);
	if(t1 != t2)
		f[t2] = t1;
	return ;
}
int main()
{
	int i,t,x,y,sum;
	scanf("%d",&t);
	while(t --)
	{
		scanf("%d%d",&n,&m);
		sum = 0;
		init();
		for(i = 1; i <= m; i ++)
		{
			scanf("%d%d",&x,&y);
			merge(x,y);
		}
		for(i = 1; i <= n; i ++)
			if(f[i] == i)
				sum ++;
		printf("%d\n",sum);
	}
	return 0;
}