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.

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.

For each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks.

2
5 3
1 2
2 3
4 5

5 1
2 5

2
4

Ignatius.L

考察知识点：

并查集，

代码如下：

#include<stdio.h>
int a[10010];
int f(int x)
{
	int r=x;
	while(r!=a[r])
	r=a[r];
	return r;
}
void merge(int x,int y)//此处是int语句的最后加上一个return 0;也是可以的 
{
	int fx=f(x);
	int fy=f(y);
	if(fx!=fy)
	{
		a[fx]=fy;
	}
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int i,n,m,x,y;
		scanf("%d%d",&n,&m);
		for(i=1;i<=n;i++)//必须从一开始 ，从零开始就WA 
		{
			a[i]=i;//初始化，并作为最后判定的条件。//将每个人的根节点初始化自身。最后统计总的根节点的个数
		}
		for(i=0;i<m;i++)
		{
			scanf("%d%d",&x,&y);
			merge(x,y);//压缩路径函数，减少跟点数量，也就是将认识的人合并
		}
		int count=0;
		for(i=1;i<=n;i++)
		{
			if(a[i]==i)
			count++;
		}
		printf("%d\n",count);
	}
	return 0;
}