POJ 1611 The Suspects (并查集求数量）

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
大意：
　　有n个人编号为0~n，m个团队，每个团队开头k表示k个人，然后是k个人的编号，编号为0的是嫌疑人，和嫌疑人在一个团队的也是嫌疑人，求嫌疑人的数量。
解题思路：
　　先输入n个学生，对fa[]初始化，然后输入m个团队的第一个人fir，后每输入一个人与fir比较，并放在一个集合，最后判断0~n个人有多少在0集合里。
并查集链接

 #include<cstdio>
 int n,m,ans,i,j,k,fir,fa[+],a;
 int find(int a)
 {
     if(a == fa[a])
     {
         return a;
     }
     else
     {
         return fa[a]=find(fa[a]);
     }
 }
 void f1(int x,int y)
 {
     int nx,ny;
     nx=find(x);
     ny=find(y);
     if(ny == )
     {
         fa[nx]=ny;
     }
     else
     {
         fa[ny]=nx;
     }
 }
 int main()
 {
     while(scanf("%d %d",&n,&m) && (m+n))
     {
         for(i =  ; i < n ; i++)
         {
             fa[i]=i;
         }
         for(i =  ; i < m ; i++)
         {
             scanf("%d %d",&k,&fir);
             for(j =  ; j < k ; j++)
             {
                 scanf("%d",&a);
                 f1(fir,a);
             }
         }
         ans=;
         for(i =  ; i < n ; i++)
         {
             if(find(i) == )
             {
                 ans++;
             }
         }
         printf("%d\n",ans);
     }
  }