文章标题 POJ 1611 ： The Suspects （并查集）

The Suspects

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
题意：有n个人，标号为0~n-1，然后有m个组，每个组有k个人，如果一个人感染了病毒，那么有这个人的组都将被感染，现知道编号为0的这个人感染了病毒，然后要求出总共有多少人感染了病毒。
分析：简单的并查集，每次将同一组的并到同一集合，然后最后将0这一集合的数目输出来
代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<vector>
#include<math.h>
#include<map>
#include<queue> 
#include<algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
int n,m;
int fa[30005];
int num[30005];
int find(int x){
    return x==fa[x]?x:fa[x]=find(fa[x]);
}
void merge (int u,int v){
    int fu=find(u);
    int fv=find(v);
    if (fu==fv) return;
    fa[fu]=fv;
    num[fv]+=num[fu];
}
int a[30005];
int main ()
{
    int k;
    while (cin>>n>>m){
        if (n==0&&m==0)break;
        for (int i=0;i<30005;i++){
            fa[i]=i;
            num[i]=1;
        }
        while (m--){
            cin>>k;
            for (int i=0;i<k;i++){
                cin>>a[i];
            }   
            for (int i=0;i<k-1;i++){//将同一组的并起来
                merge(a[i],a[i+1]);
            }
        }
        cout<<num[find(0)]<<endl;//输出0这一集合
    }       
    return 0;
}