How Many Tables

                                   How Many Tables

 

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

该题的题意是将所有直接认识或者间接认识的人安排到一桌，每桌人数不限，也就是说看看有多少组互相没有关系的人。

AC代码如下：

#include<cstdio>
int fa[1010];
int N,M;
int find(int x)
{
	return (x==fa[x])?x:find(fa[x]);
}//找到根节点//
void andd(int x,int y)
{
	int f1=find(x);
	int f2=find(y);
	if(f1!=f2)
		fa[f1]=f2;
}//合并两组有关系的圈子//
int main()
{
	int t;
	int a,b;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&N,&M);
		for(int i=1;i<=N;i++)
			fa[i]=i;//初始化//
		for(int i=1;i<=M;i++)
		{
			scanf("%d%d",&a,&b);
			andd(a,b);
		}
		int ans=0;//记录没有关系的组数//
			for(int i=1;i<=N;i++)
			{
				if(i==fa[i])
					ans++;
			}
		printf("%d\n",ans);
	}
	return 0;
}

haha