The Suspects

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.
Input
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.
Output
For each case, output the number of suspects in one line.
Sample Input
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
Sample Output
4
1
1
题意：一共有n个人（编号从1到n），m个社团，编号为1的人是感染体，和感染体在一个社团的都被怀疑为感染体，一共多少感染的？

#include<iostream>
using namespace std;
int pre[30005],num[30005],x,y,k,n,m,cnt;
int find(int x){
	if(x==pre[x]) return x;
	else return pre[x]=find(pre[x]);
}
void merge(int x,int y){
	int fx=find(x);
	int fy=find(y);
	if(fx!=fy) {
		pre[fx]=fy;
		num[fy]+=num[fx];
		/*增加了这一步，若fx的祖先是fy，
		那么fx带着他的小弟们投靠祖先fy,fy人数相应的增加 */ 
	}
}
void init(){
	cnt=0;
	for(int i=0;i<n;i++) 
	{
		pre[i]=i;
		num[i]=1;//一开始每个人是自己祖先，统领的人数只有自己，也就是 1 
	 } 
	for(int i=1;i<=m;i++){
		scanf("%d",&k);
		scanf("%d",&y);
		for(int i=1;i<k;i++){
			scanf("%d",&x);
			merge(x,y);//我这里相当于把每个社团的第一个人y当做这个社团的祖先 
		}
	}
}
int main(){
	while(scanf("%d %d",&n,&m)&&(n||m)){
		init();
		int v=find(0);
		printf("%d\n",num[v]); 
	}
}