并查集The Suspects （POJ - 1611 ）

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我是靠谱客的博主热心夏天，最近开发中收集的这篇文章主要介绍并查集The Suspects （POJ - 1611 ），觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

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：有几个团队，如果该团队存在一个人感染病毒，则该团队所有人都含有病毒，一个人可以加入多个团队，病毒携带者固定为0。求多少人可能携带病毒？

因为固定感染者为 0 号，所以将所有人统计到sum[0]中即可。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
using namespace std;
const int maxn = 3e4+5;
int fa[maxn],sum[maxn],n,m;
int findfa(int x)
{
    return x == fa[x]?x:fa[x] = findfa(fa[x]);
}
void Union(int a,int b) //a<b;
{
    int x = findfa(a),y = findfa(b); // x < y;
    if( x > y)
        swap(x,y); //确定所有的父节点为 0号 节点
    if(x!=y)
    {
        fa[y] = x;
        sum[x] += sum[y];
    }
}
int main()
{
    while(scanf("%d%d",&n,&m) &&(n!=0||m!=0) )
    {
        for(int i = 0; i < n; i++)
        {
             sum[i] = 1;
             fa[i] = i;
        }
        int a,b,k;
        while(m--)
        {
            cin>>a>>b;
            a--;
            while(a--)
            {
                cin>>k;
                Union(b,k);
            }
        }
        cout<<sum[0]<<endl;
    }
    return 0;
}