poj 1611~并查集基础

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

#include<iostream>
#include<cstdio>

int pre[30050];

void init(int n)
{
    for(int i=0;i<n;i++)
        pre[i]=i;
}

int find(int x)
{
   if(x==pre[x]) return x;
   int temp=find(pre[x]);
   pre[x]=temp;
   return temp;
}

void Union(int u,int v)
{
    int p=find(u);
    int q=find(v);
    if(p!=q)
        pre[p]=q;
}

int main()
{
    int n,m;
    while(~scanf("%d %d",&n,&m)&&(n||m))
    {
        init(n);
        int a,b,c,d;
        int x;
        while(m--)
        {
            scanf("%d",&x);
            if(x==1) {scanf("%d",&b);continue;}
            else
            {
                scanf("%d",&a);
                x--;
                while(x--)
                {
                    scanf("%d",&b);
                    Union(a,b);
                    a=b;
                }

            }

        }
        int cnt=0;
        int t=find(0);
        for(int i=0;i<n;i++)
        {
            if(find(i)==t) cnt++;  //期初wa是因为这里、、这种路径压缩的写法可能造成不用find函数无法完成对树的非根节点的根节点的更新

        }

        printf("%d\n",cnt);
    }
    return 0;
}

　　题目大意是说n个人，m组，只要一个组中一个人得了sars，一个组中其他人也被怀疑、并查集不解释，默认编号为0的人得了sars