The Suspects(POJ--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.

题意：多组输入，每组输入n和m分别表示有n个学生和m个小组，每个小组有k个学生。学生编号从0到n-1，规定每组数据0号学生肯定是传染者，只要一个小组有一个传染者则这整个小组都是传染者，一个学生可以在多个小组，求总共有多少个传染者

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

#include <cstdio>#include <queue>#include <cstring>#define MAX 100002using namespace std;int pre[31000];int Find(int x){    int r,j,k;    r=x;    while(r!=pre[r])        r=pre[r];    k=x;    while(k!=r)    {        j=pre[k];        pre[k]=r;        k=j;    }    return r;}int main(){    //freopen("oo.text","r",stdin);    int n,m,k,x,y,fx,fy;    while(~scanf("%d %d",&n,&m))    {        if(n==0&&m==0)            break;        for(int i=0; i<n; i++)            pre[i]=i;        while(m--)        {            scanf("%d",&k);            scanf("%d",&x);            fx=Find(x);            for(int j=1; j<=k-1; j++)            {                scanf("%d",&y);                fy=Find(y);                if(fx!=fy)                {                    if(fy==0)                        pre[fx]=fy;                    else                        pre[fy]=fx;                }            }        }        for(int i=0;i<n;i++)          //更新所有点        pre[i]=Find(pre[i]);        int sum=0;        for(int i=0; i<n; i++)        {            //printf("******%d %d\n",i,pre[i]);            if(pre[i]==0)                sum++;        }        printf("%d\n",sum);    }    return 0;}<strong></strong>