The Suspects

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

本题大意：非典爆发，告诉你一共有n个人，编号分别为从0到n-1，
最开始只有编号为0的人被感染。有m个团体，
团体内有一个人感染其他人就都有可能感染，
接下来的m行为各个团体人员，k为这个团体的人数。

题目思路：题目实际上就是让我们把每个人之间有关系的标出来，
最后找和编号为0的同学有关系的人有多少个（包括0本身）。
用并查集把每个团的人并到一个集合。

 

这道题就是简单的运用并查集的模板。

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int N=30000+20;
int fa[N];
int find(int x)
{
    if(fa[x]==x) return x;
    return find(fa[x]);
}
void join(int x,int y)
{
    int fx,fy;
    fx=find(x),fy=find(y);
    if(fx!=fy)
        fa[fy]=fx;
}
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m)&&(n+m))
    {
        for(int i=0;i<=n;i++) fa[i]=i;
        int k,x,y;
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&k,&x);
            for(int j=1;j<k;j++)
            {
                scanf("%d",&y);
                join(x,y);
            }
        }
        int cnt=0;
        for(int i=0;i<n;i++)
            if(find(i)==find(0))
            cnt++;
        printf("%d\n",cnt);
    }
    return 0;
}