POJ 1611 The Suspects（并查集）

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

题目大意：某个学校里的人感染了病毒，现在要把他们隔离起来。病毒会在不同的群体之间传染，，如果一个群体中有一个人感染了病毒，那么这一个群体中的所有人都会被感染。现在要求出有多少人感染了病毒，以便隔离他们。默认0号同学始终感染病毒。

解题思路：将感染了病毒的人放在一个集合中，其他人放在其他的集合中，所以要用到并查集。注意输入的格式。

代码如下：

#include <algorithm>
#include <cctype>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#define EPS 1e-6
#define INF INT_MAX / 10
#define LL long long
#define MOD 100000000
#define PI acos(-1.0)

const int maxn = 30005;
int par[maxn],rank[maxn];

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

int find(int x)
{
    return par[x] == x ? x : par[x] = find(par[x]);
}

void unite(int x,int y)
{
    x = find(x);
    y = find(y);
    if(x == y)
        return ;
    if(rank[x] < rank[y]){
        par[x] = y;
    }
    else{
        par[y] = x;
        if(rank[x] == rank[y])
            rank[x]++;
    }
}

int main()
{
    int n,m;
    while(scanf("%d %d",&n,&m) != EOF && (n || m)){
        init(n);
        while(m--){
            int k,a,b;
            scanf("%d",&k);
            if(k > 0){
                scanf("%d",&a);
                for(int i = 1;i < k;i++){
                    scanf("%d",&b);
                    unite(a,b);
                }
            }

        }
        int ans = 0;
        int f = find(0);
        for(int i = 0;i < n;i++){
            if(find(i) == f)
                ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}