poj 1611 The Suspects

                                               The Suspects

Time Limit: 1000MS Memory Limit: 20000KTotal Submissions: 37732 Accepted: 18302

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

简要题意：

有n个同学，m个小组，其中每个同学有一个唯一编号（0-n-1），0号同学默认为感染者，和0号同学处在同一个小组内的同学均视为感染者，而且同一个人可以加入多个小组，若组内有感染者则整个小组均视为感染者，求感染者的人数。

分析：

并查集的应用，感染者合并为一个集合，最后求该集合元素的个数。

#include <iostream>#include <stdio.h>#include <algorithm>#include <string.h>#include <math.h>using namespace std;const int MAXN=30000+5;int pre[MAXN];  //记录前驱节点int findRoot(int x)  //求某个元素的根节点{    while(pre[x]!=x)        x=pre[x];    return x;}void merge(int x,int y)  //合并两个元素{    int xRoot=findRoot(x);    int yRoot=findRoot(y);    if( xRoot!=yRoot )        pre[xRoot]=yRoot;}int main(){    int n,m;    while(~scanf("%d %d",&n,&m))    {        if(m==0 && n==0)            break;        int i,j;        for(i=0;i<n;++i)        pre[i]=i;        for(i=0;i<m;++i)        {            int a;            scanf("%d",&a);            int b[MAXN];            for(j=0;j<a;++j)            {                int c;                scanf("%d",&c);                b[j]=c;                merge(b[0],b[j]);            }        }        int mycount=0;        for(i=0;i<n;++i)        {            if(findRoot(i)==findRoot(0))  //求与0号同学根节点相同的人数，即为答案                mycount++;        }        printf("%d\n",mycount);    }    return 0;}