The Suspects(并查集维护根节点信息)

The Suspects

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：总共有n个学生编号0～n-1，总共m个社团，现在0号学生是可能感染病菌的学生，那么和他在一个社团的学生需要
    被隔离，这些学生如果还在其他社团的话，那么那个社团的所有学生也要被隔离，现在问你需要隔离多少学生

初步思路：并查集，用一个数组来维护一棵树的节点数量，所有只需要输出0所在的树的节点数量就行了
*/
#include <iostream>
#include <stdio.h>
using namespace std;
int n,m;
int a,last;
int k;
int bin[30005];
int num[30005];
int findx(int x){
    int tmp=x;
    while(x!=bin[x]){
        x=bin[x];
    }
    bin[tmp]=x;//路径压缩
    return x;
}
int bindx(int x,int y){
    int fx=findx(x);
    int fy=findx(y);
    if(fx!=fy){
        num[fy]+=num[fx];//这两个点相联合了，那么根节点的信息要加到一块
        bin[fx]=fy;
        return 1;
    }
    return 0;
}
void init(){
    for(int i=0;i<=n;i++){
        bin[i]=i;
        num[i]=1;
    }
}
int main(){
    // freopen("in.txt","r",stdin);
    while(scanf("%d%d",&n,&m)!=EOF&&(n+m)){
        init();
        while(m--){
            scanf("%d",&k);
            for(int i=0;i<k;i++){
                scanf("%d",&a);
                if(i){
                    bindx(a,last);
                }
                last=a;
            }
        }
        printf("%d\n",num[findx(0)]);
    }
    return 0;
}