The Suspects POJ

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 42 1 25 10 13 11 12 142 0 12 99 2200 21 55 1 2 3 4 51 00 0

Sample Output

41
1
思路：题目要求含0树的节点个数。把每个小组的人都进行关联，先取出每组第一个人，把第一个人和剩下k-1个人unite一下，就把关联的人全部生成一个树。这样就会生成>=1个树。其中用cnt数组记录每次unite后根节点的节点数。最后输出0对应的根节点的cnt数，OVER！
#include <stdio.h>int parent[30005];   //用来记录父亲节点int cnt   [30005];   // 记录根节点的节点数int n;int group;int find(int x){    return (parent[x]==x)?x:parent[x]=find(parent[x]);}void unite(int x,int y){    x=find(x);    y=find(y);    if(x!=y)    {        parent[y]=x;        cnt[x]+=cnt[y];  //这里是解题的关键，我还得再思考一下。    }}void init(){    for(int i=0;i<n;i++)    {        parent[i]=i;        cnt[i]=1;    }}int main(void){    while((scanf("%d%d",&n,&group))==2)    {        if(n==0 && group == 0) return 0;        init();        for(int i=0;i<group;i++)        {            int num;            scanf("%d",&num);            int temp;            scanf("%d",&temp);            for(int j=2;j<=num;j++)            {                int temp1;                scanf("%d",&temp1);                unite(temp,temp1);            }        }        printf("%d\n",cnt[find(0)]);    }}