POJ 1611 The Suspects

The Suspects

Time Limit: 1000MS
Memory Limit: 20000KTotal Submissions: 22220
Accepted: 10808

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

Source

Asia Kaohsiung 2003

题目大意  :一共有n个学生编号0～n-1，给出m组，规定0号学生有传染病，凡是与其在一组的都会被传染
                 要求出最后一共有多少学生会得病

题目分析  :裸的并查集，按秩合并

#include <cstdio>int const MAX = 30000 + 10;int rank[MAX];  //秩，表示当前集合中的人数int father[MAX]; void UFset(int n)  //初始化{    for(int i = 0; i < n; i++)    {        father[i] = i;        rank[i] = 1;    }}int Find(int x) //查找，路经压缩{    return x == father[x] ? x : father[x] = Find(father[x]);}void Union(int R1, int R2)   //按秩合并{    int r1 = Find(R1);    int r2 = Find(R2);    if(r1 == r2)        return;    if(rank[r1] <= rank[r2])   //把秩大的作为父亲节点    {        father[r1] = r2;        rank[r2] = rank[r1] + rank[r2];    }    else    {        father[r2] = r1;        rank[r1] = rank[r1] + rank[r2];    }}int main(){    int n, m;    while(scanf("%d%d",&n, &m) != EOF && (m + n))    {        UFset(n);        int group, first, next;        for(int j = 0; j < m; j++)        {            scanf("%d%d", &group, &first);            for(int i = 1; i < group; i++)            {                scanf("%d", &next);                Union(first,next);            }        }        printf("%d\n",rank[Find(father[0])]);  //输出0所属的根节点的秩极为0所在集合的人数    }}