POJ1611_The Suspects

The Suspects

Time Limit: 1000MS Memory Limit: 20000KTotal Submissions: 22366 Accepted: 10875

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：

n个学生分属m个团体，(0 < n <= 30000 , 0 <= m <= 500) 一个学生可以属于多个团体。一个学生疑似患病，则它所属的整个团体都疑似患病。已知0号学生疑似患病，以及每个团体都由哪些学生构成，求一共多少个学生疑似患病。

 

思路：并查集的最基本应用。把所有可疑的都并一块。其中，要用一个total数组，total[find(a)]是a所在的
集合的人数。最后输出0号节点所在集合的感染总数，也就是0号节点根节点的感染总数。

 

代码：

# include<stdio.h># include<string.h>const int MAXN = 30010;int father[MAXN],total[MAXN];int find(int a)//获取a的根,并把a的父节点改为根{    if(a != father[a])        father[a] = find(father[a]);    return father[a];}int main(){    int n, m, k, a, b;    while(scanf("%d %d", &n, &m) && (n||m) )    {        memset(father,0,sizeof(father));        memset(total,0,sizeof(total));        for(int i = 0; i < n; i++)        {            father[i] = i;            total[i] = 1;        }        for(int i = 0; i < m; i++)        {            scanf("%d", &k);            scanf("%d", &a);            for(int j = 1; j < k; j++)            {                scanf("%d", &b);                a = find(a);                b = find(b);                if(a!=b)                {                    father[b] = a;                    total[a] += total[b];                }                            }        }        printf("%d\n",total[find(0)]);//必须输出的是0的根节点的总数，因为可能还没进行路径压缩。    }    return 0;}