【POJ 1611 The Suspects】+ 并查集

The Suspects
Time Limit: 1000MS Memory Limit: 20000K
Total Submissions: 35581 Accepted: 17269

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 4
2 1 2
5 10 13 11 12 14
2 0 1
2 99 2
200 2
1 5
5 1 2 3 4 5
1 0
0 0

Sample Output

4
1
1

题意 : 有 N 个同学,M 个社团，兹定0号为嫌疑人，任何和嫌疑人有关系的人都化为嫌疑人，每个社团的第一个数 K 为社团人数，接下来为 K 名社团成员的～

思路 ： 把所有有关系的人并在一起，输出 嫌疑人0所在的集合人数

AC代码：

#include<cstdio>#include<cstring>using namespace std;int f[30010],s[30010];int bc(int x){   int p = x;   while(x != f[x]) // 寻找父亲节点      x = f[x];   while(p != x){ // 路径压缩      int t = f[p];      f[p] = x;      p = t;   }   return x;}void un(int x,int y){    int fx = bc(x),fy = bc(y);    if(fx != fy){        f[fy] = fx;        s[fx] += s[fy];    }}int main(){    int N,M,K,a,b;    while(scanf("%d %d",&N,&M) != EOF && (N || M)){        for(int i = 0 ; i <= N; i++)            f[i] = i,s[i] = 1;        while(M--){            scanf("%d %d",&K,&a);            while(--K){                scanf("%d",&b);                un(a,b); // 把有关联的并在一起            }        }        printf("%d\n",s[bc(0)]);    }    return 0;}