POJ 1611 The Suspects 【并查集】

题面：

evere 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

题目大意：

现在有n个人，编号为0-n。有m个团体，每个团体k个人。已知0号是有嫌疑，所以与0号是一个团体的都有嫌疑。问有多少个有嫌疑的人。

大致思路：

很明显的思路，把所有提到的人利用并查集进行集合判断。然后查找和0同一集合的人就行了。现在存在一种情况：提到的人数比较少，而n又比较大。这个时候从1-n扫一遍就很浪费时间。所以利用set，将提到的人的编号保存起来。然后只需要找提到的人就行了。

代码：

#include<iostream>
#include<algorithm>
#include<set>
#include<cstdlib>
using namespace std;
const int maxn=3e4+10;
int father[maxn];
int Find(int x)
{
    if(x!=father[x])
        father[x]=Find(father[x]);
    return father[x];
}
void Union(int x,int y)
{
    x=Find(x);
    y=Find(y);
    if(x!=y)
        father[x]=y;
    return ;
}
int main()
{
    ios::sync_with_stdio(false);
    //freopen("in.txt","r",stdin);
    set<int> s;
    int n,m,k,cnt,x,y;
    while(cin>>n>>m&&(n||m))
    {
        s.clear();
        cnt=1;
        for(int i=0;i<n;++i)
            father[i]=i;
        for(int i=0;i<m;++i){
            cin>>k>>x;
            s.insert(x);//将读到的插入set
            for(int i=1;i<k;++i){
                cin>>y;
                s.insert(y);
                Union(x,y);
            }
        }
        int key=Find(0);
        set<int>::iterator it;
        for(it=s.begin();it!=s.end();++it){//将set中的元素扫一遍
            if(key==Find(*it)&&(*it))//父节点相同且元素不为0，防止重复计算
                cnt++;
        }
        cout<<cnt<<endl;
    }
    return 0;
}