博客
关于我
【 POJ - 1611 】 C - The Suspects(简单并查集)求集合中元素个数
阅读量:275 次
发布时间:2019-03-01

本文共 2177 字,大约阅读时间需要 7 分钟。

题目:

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:

学生0都被认为是可疑者 , 求与0在同一集合的人数(包括0)


代码:

#include 
#include
#include
using namespace std;const int maxn=3e4+50;int father[maxn],n,m,c,f,x;int find(int x){ //找根,路径压缩 return x==father[x]?x:father[x]=find(father[x]);}void Union(int a,int b) //合并{ int fa=find(a); int fb=find(b); if(fa!=fb) father[fa]=fb;}void init() //初始化{ for(int i=0;i

转载地址:http://ctao.baihongyu.com/

你可能感兴趣的文章
MySQL DELETE 表别名问题
查看>>
MySQL Error Handling in Stored Procedures---转载
查看>>
MVC 区域功能
查看>>
MySQL FEDERATED 提示
查看>>
mysql generic安装_MySQL 5.6 Generic Binary安装与配置_MySQL
查看>>
Mysql group by
查看>>
MySQL I 有福啦,窗口函数大大提高了取数的效率!
查看>>
mysql id自动增长 初始值 Mysql重置auto_increment初始值
查看>>
MySQL in 太多过慢的 3 种解决方案
查看>>
MySQL InnoDB 三大文件日志,看完秒懂
查看>>
Mysql InnoDB 数据更新导致锁表
查看>>
Mysql Innodb 锁机制
查看>>
MySQL InnoDB中意向锁的作用及原理探
查看>>
MySQL InnoDB事务隔离级别与锁机制深入解析
查看>>
Mysql InnoDB存储引擎 —— 数据页
查看>>
Mysql InnoDB存储引擎中的checkpoint技术
查看>>
Mysql InnoDB存储引擎中缓冲池Buffer Pool、Redo Log、Bin Log、Undo Log、Channge Buffer
查看>>
MySQL InnoDB引擎的锁机制详解
查看>>
Mysql INNODB引擎行锁的3种算法 Record Lock Next-Key Lock Grap Lock
查看>>
mysql InnoDB数据存储引擎 的B+树索引原理
查看>>