POJ-1611 ----The Suspects

The Suspects

Time Limit: 1000MS Memory Limit: 20000KTotal Submissions: 39931 Accepted: 19315

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

题意：在SARS病毒期间学校要查可能感染这种病毒的人数，学院有n个学生，m个学生团体，一个学生可以加入多个学生团体，有学生在学生团体的名单，如果一名感染学生在这个学生团体这个团体的所有人都会成为病毒携带嫌疑人，找到这些嫌疑人的总数。无论哪种情况否认为0已经感染病毒。

思路：判断有多少个父节点与0相同的，使用并查集。

代码：

#include<stdio.h>#include<string.h>int f[30300],m,n;//注意数组大小void init(){    int i;    for(i=0;i<n;i++)        f[i]=i;}int findx(int x){    if(f[x]==x)        return x;    else        return f[x]=findx(f[x]);}void merge1(int x,int y){    int tx,ty;    tx=findx(x);    ty=findx(y);    if(tx!=ty)    {        f[ty]=tx;    }}int main(){    while(scanf("%d%d",&n,&m)&&m+n)    {        int k,a,b,sum=1;        init();        for(int i=1 ; i<=m ; i++)        {            scanf("%d",&k);            scanf("%d",&a);            for(int j=1;j<k;j++)            {                scanf("%d",&b);                merge1(a,b)//a,b是否相连            }        }        for(int i=1;i<n;i++)        {            if(findx(i)==findx(0))//判断是否和0的父节点相同                sum++;        }        printf("%d\n",sum);    }    return 0;}