POJ 1611 并查集入门

http://poj.org/problem?id=1611

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 42 1 25 10 13 11 12 142 0 12 99 2200 21 55 1 2 3 4 51 00 0

Sample Output

411

并查集讲解http://dongxicheng.org/structure/union-find-set/

本题代码：

#include <stdio.h>#include <iostream>#include <string.h>using namespace std;int n,m,k,fa[30010],num[30010],x,y;void initset()//初始状态树的所有父亲节点都是其本身，该树的所含元素为1；{    for(int i=0;i<n;i++)    {        fa[i]=i;        num[i]=1;    }}int find(int x)//寻找该点的根节点{    if(x==fa[x])        return x;    return fa[x]=find(fa[x]);}void un(int x,int y)//若二者的根节点不同，则连到一起作为一个集合，该集合所含元素为先前二者之和{    x=find(x);    y=find(y);    if(x!=y)    {        fa[y]=x;        num[x]+=num[y];    }}int main(){    while(~scanf("%d%d",&n,&m))    {        if(n==0&&m==0)             break;        initset();        for(int i=0;i<m;i++)        {            scanf("%d",&k);               scanf("%d",&x);//输入一组数该组数一定在同一棵树上            for(int j=1;j<k;j++)            {                scanf("%d",&y);                un(x,y);            }        }        printf("%d\n",num[find(0)]);//树出和0有共同节点的树（即与0同一棵树）里的元素的个数//此处用num[a[0]]就错，我不知道为什么    }    return 0;}