poj1611 简单并查集

The Suspects

Time Limit: 1000MS		Memory Limit: 20000K
Total Submissions: 32781		Accepted: 15902

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
解析见代码：

/*
题目大意：有n个人，编号0~n-1,其中0号可能携带有病毒
和他在一个团体里的人也可能有病毒，同时也是病毒传染者，
问你一共有多少个人可能携带病毒
思路分析：简单并查集，将元素合并为若干个集合，注意在merge
过程中同时维护该集合中的人数，最后输出0号所在的集合的人数即可
*/
#include <iostream>
#include <cstdio>
#include<cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn=30000+100;
int father[maxn];
int total[maxn];//合并集合的时候需要将两个集合中的元素数目同时进行合并
int ans;
int n,m;
int findroot(int x)
{
    return (x==father[x])?x:father[x]=findroot(father[x]);
}
void merge(int a,int b)
{
    int x=findroot(a);
    int y=findroot(b);
    if(x==y) return;//两者在同一个集合里
    total[x]+=total[y];
    father[y]=x;
}
int main()
{
    while(scanf("%d%d",&n,&m)&&(n||m))
    {
        for(int i=0;i<n;i++)//并查集初始化
        {
            father[i]=i;
            total[i]=1;
        }
        int k;
        int a,b;;
        while(m--)
        {
            scanf("%d",&k);
            scanf("%d",&a);
            for(int i=1;i<k;i++)
            {
                scanf("%d",&b);
                merge(a,b);
            }
        }
        printf("%d
",total[findroot(father[0])]);//此处findroot(father[0])不能写作
    }
}