POJ 1611 The Suspects

The Suspects

Time Limit: 1000MS		Memory Limit: 20000K
Total Submissions: 24044		Accepted: 11741

Description

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

题目大意：有n个人，m组。0是病毒携带者，和0一组的人被怀疑感染病毒，和被怀疑的人一组的人也被怀疑。求所有被怀疑者的个数。

解题思路：并查集。将0作为祖先，祖先为0的都是被怀疑者。

代码如下：

#include <cstdio>
int const maxn=30003;
int father[maxn];
int find(int x)
{
	return x==father[x]?x:father[x]=find(father[x]);
}
void Union(int a,int b)
{
	int a1=find(a);
	int b1=find(b);
	if(a1>b1)
	father[a1]=b1;
	else
		father[b1]=a1;
}
int main()
{
	int n,m,cnt,i,j,num,f,x;
	while(scanf("%d%d",&n,&m)!=EOF&&(n||m))
	{
		cnt=0;
		for(i=0;i<n;i++)
			father[i]=i;
		for(i=0;i<m;i++)
		{
			scanf("%d%d",&num,&f);
			for(j=0;j<num-1;j++)
			{
				scanf("%d",&x);
				Union(f,x);
			}
		}
		for(i=0;i<n;i++)
			if(find(i)==0)
				cnt++;
		printf("%d\n", cnt);
	}
	return 0;
}