并查集 Ⅱ

[poj 1611] (http://poj.org/problem?id=1611)
题目描述：

The Suspects（嫌疑犯）
Time Limit: 1000MS Memory Limit: 20000K
Total Submissions: 27293 Accepted: 13318
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
言简意赅：
此题是要计算可能受感染者的数量，已知共有n人和m个团体，处于同一个团体中的人之间能产生传染的可能，且已知0号同学是病毒携带者；则此题可用并查集的方法来解决，要特别注意输入时的处理，可以先将每个团体中的人连起来，保证根节点为团体中第一个人，并设置一个num数组记录下来每个团体中的人数，最后将含有0的个数输出即可

代码实现：

#include <iostream>
#include <stdio.h>

using namespace std;

const int N=30005;
int father[N];
int num[N];//num[]存储节点所在集合元素的个数，father[]存储节点的父亲节点

void make_set(int x)//初始化集合
{
    father[x]=-1;
    num[x]=1;
}

int find_set(int x)//查找x元素所在的集合，并返回根节点
{
    int r,tmp;
    r=x;//记录下待查找值x
    while(father[r]!=-1)
    {
        r=father[r];//找到根节点
    }
    while(x!=r)//压缩路径，将路径上所有x的子孙都连接到r上
    {
        tmp=father[x];
        father[x]=r;
        x=tmp;
    }
    return x;
}
void union_set(int a,int b)
{
    a=find_set(a);
    b=find_set(b);
    if(a==b)//若两个元素在同一个集合，不需要合并
    return;
    if(num[a]<=num[b])//将小集合合并到大集合中
    {
        father[a]=b;
        num[b]+=num[a];
    }
    else
    {
        father[b]=a;
        num[a]+=num[b];
    }
}

int main()
{
    int n,m,k,a,b;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n==0&&m==0)break;//if(n+m==0)break;
        for(int i=0;i<n;i++)
        {
            make_set(i);
        }
        for(int i=0;i<m;i++)
        {
            scanf("%d",&k);
            scanf("%d",&a);
            for(int j=1;j<k;j++)
            {
                scanf("%d",&b);
                union_set(a,b);
            }
        }
        printf("%d\n",num[find_set(0)]);
    }
    return 0;
}