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中对于数据库的基本操作
查看>>
Mysql中常用函数的使用示例
查看>>
MySql中怎样使用case-when实现判断查询结果返回
查看>>
Mysql中怎样使用update更新某列的数据减去指定值
查看>>
Mysql中怎样设置指定ip远程访问连接
查看>>
mysql中数据表的基本操作很难嘛,由这个实验来带你从头走一遍
查看>>
Mysql中文乱码问题完美解决方案
查看>>
mysql中的 +号 和 CONCAT(str1,str2,...)
查看>>
Mysql中的 IFNULL 函数的详解
查看>>
mysql中的collate关键字是什么意思?
查看>>
MySql中的concat()相关函数
查看>>
mysql中的concat函数,concat_ws函数,concat_group函数之间的区别
查看>>
MySQL中的count函数
查看>>
MySQL中的DB、DBMS、SQL
查看>>
MySQL中的DECIMAL类型:MYSQL_TYPE_DECIMAL与MYSQL_TYPE_NEWDECIMAL详解
查看>>
MySQL中的GROUP_CONCAT()函数详解与实战应用
查看>>
MySQL中的IO问题分析与优化
查看>>
MySQL中的ON DUPLICATE KEY UPDATE详解与应用
查看>>
mysql中的rbs,SharePoint RBS:即使启用了RBS,内容数据库也在不断增长
查看>>
mysql中的undo log、redo log 、binlog大致概要
查看>>