Severe acute respiratory syndrome (SARS), an atypical pneumonia of unknown aetiology, was recognized as a global threat in mid-March 2003. To minimize transmission to others, the best strategy is to separate the suspects from others.
In the Not-Spreading-Your-Sickness University (NSYSU), there are many student groups. Students in the same group intercommunicate with each other frequently, and a student may join several groups. To prevent the possible transmissions of SARS, the NSYSU collects the member lists of all student groups, and makes the following rule in their standard operation procedure (SOP).
Once a member in a group is a suspect, all members in the group are suspects.
However, they find that it is not easy to identify all the suspects when a student is recognized as a suspect. Your job is to write a program which finds all the suspects.

The input file contains several cases. Each test case begins with two integers n and m in a line, where n is the number of students, and m is the number of groups. You may assume that 0 < n <= 30000 and 0 <= m <= 500. Every student is numbered by a unique integer between 0 and n−1, and initially student 0 is recognized as a suspect in all the cases. This line is followed by m member lists of the groups, one line per group. Each line begins with an integer k by itself representing the number of members in the group. Following the number of members, there are k integers representing the students in this group. All the integers in a line are separated by at least one space.
A case with n = 0 and m = 0 indicates the end of the input, and need not be processed.

For each case, output the number of suspects in one line.

100 4
2 1 2
5 10 13 11 12 14
2 0 1
2 99 2
200 2
1 5
5 1 2 3 4 5
1 0
0 0

4
1
1

题解：

题意大概就是，一个学生可以加入很多个不同的组织，学生0得了SARS，那么，和0学生在一个协会的同学以及和0学生在一个协会的同学的同学都被列为被传染的嫌疑人，输出一共多少人被视为嫌疑人。

接下来就很简单明了了，并查集找出0学生以及和0学生有间接关系的所有同学在一个庞大的组织里，所以，建立起这个组织之后，就通过0同学找到组织老大，然后再通过组织老大依次判断谁的组织老大是这个老大，就sum++；

代码：

#include <stdio.h>
#include <algorithm>

using namespace std;

int pre[60000];

int find(int x)
{
	int r=x;
	while(pre[r]!=r)
	r=pre[r];
	int s;
	while(pre[x]!=r)
	{
		s=pre[x];
		pre[x]=r;
		x=s;
	}
	return r;
}

void join(int x,int y)
{
	int fx=find(x),fy=find(y);
	if(fx!=fy)
		pre[fx]=fy;
}

int main()
{
	int n,m,k;
	int x,y;
	while(~scanf("%d%d",&n,&m))
	{
		int sum=0;
		int ans=1;
		int a;
		if(n==0&&m==0) break;
		for(int i=0;i<n;i++)
		{
			pre[i]=i;
		}
		for(int i=0;i<m;i++)
		{
			scanf("%d",&k);
			if(k==0) continue;
			scanf("%d",&a);
			for(int i=1;i<k;i++)
			{
				scanf("%d",&x);
				join(a,x);	
			}	
		}
		sum=find(0);
		for(int i=1;i<n;i++)
			if(sum==find(i))
				ans++;
		printf("%d\n",ans);	
	}	
	return 0;
}