【北大夏令营笔记-并查集】poj1611-The Suspects

The Suspects
Time Limit: 1000MS Memory Limit: 20000K
Total Submissions: 21517 Accepted: 10422
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
Source

Asia Kaohsiung 2003

//题意，有一些学生，加入了一些社团，包括0。一旦0患病，和他有关的人（同一团体）都是嫌疑对象。

问有多少嫌疑对象

//思路：并查集，将所有与0有关的人都放在同一集合，并把与0有关
的人所在的社团的所有人数统计

AC代码：

#include<stdio.h>
#define MAX 30000
int parent[MAX+10];
int total[MAX+10];
//total[Get_Root(a)]是a所在社团的人数
int Get_Root(int x)
{
    //获取a的根，并把a父节点改为根
    if(parent[x]!=x)
      parent[x]=Get_Root(parent[x]);
    return parent[x];
}
void Merge(int x,int y)
{
     x=Get_Root(x);
     y=Get_Root(y);
     if(x==y)
     return;
     parent[y]=x;
     total[x]+=total[y];
}
int main()
{
    int i,j,n,m,k,x,y;
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        if(n==0&&m==0)
        break;
        for(i=0;i<n;i++)
        {
           parent[i]=i;
           total[i]=1;
        }
        for(i=0;i<m;i++)
        {
           scanf("%d",&x);
           scanf("%d",&k);
           x--;
           while(x--)
           {
              scanf("%d",&y);
              Merge(k,y);
           }
        }
        printf("%d\n",total[Get_Root(0)]);
    }
    return 0;
}