The Suspects

Time Limit: 1000MS   Memory Limit: 20000K
Total Submissions: 39211   Accepted: 18981

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

Source

 
 //2017-07-19
#include <iostream> using namespace std; const int N = ;
int n, m, fa[N]; void init(){
for(int i = ; i < N; i++)
fa[i] = i;
} int getfa(int x){
if(x == fa[x])return x;
else return fa[x] = getfa(fa[x]);
} void merge0(int a, int b){
int af = getfa(a);
int bf = getfa(b);
if(af != bf)
fa[bf] = af;
} int main(){
while(cin>>n>>m){
if(!n && !m)break;
init();
int k, a, b;
while(m--){
cin>>k>>a;
for(int i = ; i < k-; i++){
cin>>b;
merge0(a, b);
}
}
int ans = ;
for(int i = ; i < n; i++){
if(getfa(i) == fa[])
ans++;
}
cout<<ans<<endl;
} return ;
}

POJ1611(KB2-B)的更多相关文章

  1. 并查集 poj1611&poj2492

    poj1611 简单题 代码中id记录父节点,sz记录子树规模.一个集合为一棵树. #include <iostream> #include <cstdio> using na ...

  2. poj1611 The Suspects(并查集)

    题目链接 http://poj.org/problem?id=1611 题意 有n个学生,编号0~n-1,m个社团,每个社团有k个学生,如果社团里有1个学生是SARS的疑似患者,则该社团所有人都要被隔 ...

  3. poj1611(并查集)

    题目链接:http://poj.org/problem?id=1611 题意: SARS(非典型肺炎)传播得非常厉害,其中最有效的办法是隔离那些患病.和患病者接触的人.现在有几个学习小组,每小组有几个 ...

  4. poj1611 并查集 (路径不压缩)

    http://poj.org/problem?id=1611 题目大意: 有一个学校,有N个学生,编号为0-N-1,现在0号学生感染了非典,凡是和0在一个社团的人就会感染,并且这些人如果还参加了别的社 ...

  5. poj1611(感染病患者)

    The Suspects Time Limit: 1000MS   Memory Limit: 20000K Total Submissions: 24587   Accepted: 12046 De ...

  6. poj1611 带权并查集

    题意:病毒蔓延,现在有 n 个人,其中 0 号被认为可能感染,然后给出多个社交圈,如果某个社交圈里有人被认为可能被感染,那么所有这个社交圈里的人都被认为可能被感染,现在问有多少人可能被感染. 带权并查 ...

  7. poj-2236 Wireless Network &&poj-1611 The Suspects && poj-2524 Ubiquitous Religions (基础并查集)

    http://poj.org/problem?id=2236 由于发生了地震,有关组织组把一圈电脑一个无线网,但是由于余震的破坏,所有的电脑都被损坏,随着电脑一个个被修好,无线网也逐步恢复工作,但是由 ...

  8. 暑假集训(2)第二弹 ----- The Suspects(POJ1611)

    B - The Suspects Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:20000KB   ...

  9. poj1611 并查集

    题目链接:http://poj.org/problem?id=1611 #include <cstdio> #include <cmath> #include <algo ...

  10. poj1611 简单并查集

    The Suspects Time Limit: 1000MS   Memory Limit: 20000K Total Submissions: 32781   Accepted: 15902 De ...

随机推荐

  1. urllib2 的get请求与post请求

    urllib2默认只支持HTTP/HTTPS的GET和POST方法 urllib.urlencode() urllib和urllib2都是接受URL请求的相关参数,但是提供了不同的功能.两个最显著的不 ...

  2. 前端基础-html 介绍和head标签 ( 1 )

    主要内容 web标准 浏览器介绍 开发工具介绍 HTML介绍 HTML颜色介绍 HTML规范 HTML结构详解 写在前面: 前端               后端 C(client)        S ...

  3. css居中小结

    从css入门就开始接触,无所不在的,一直备受争议的居中问题. css居中分为水平居中和垂直居中,水平居中方式也较为常见和统一,垂直居中的方法就千奇百怪了. 博客原文地址:Claiyre的个人博客 ht ...

  4. POJ 2665

    #include<iostream> #include<stdio.h> using namespace std; int main() { //freopen("a ...

  5. 03-01 Java运算符

    (1)算术运算符 A:+,-,*,/,%,++,-- B:+的用法 a:加法 b:正号 c:字符串连接符 C:/和%的区别 数据做除法操作的时候,/取得是商,%取得是余数 D:++和--的用法 a:他 ...

  6. [Umbraco] 创建第一个页面

    如何创建一个页面,很简单. 进入settings,首先右键点击“Document Types”, 在出现的菜单点击"Create",在弹出的窗口中 Master Document ...

  7. Deep Learning (中文版&英文版)

    Bengio Yoshua,Ian J. Goodfellow 和 Aaron Courville共同撰写的<深度学习>(Deep Learning)是一本为了帮助学生及从业者入门机器学习 ...

  8. centos 7 安装 mysql 5.7

    1.环境 Centos 7 2.下载 官方网站https://dev.mysql.com/downloads/mysql/5.7.html#downloads ,选择要下载的版本,centos 7 同 ...

  9. vector源码1(参考STL源码--侯捷):源码

    vector源码1(参考STL源码--侯捷) vector源码2(参考STL源码--侯捷) vector源码(参考STL源码--侯捷)-----空间分配导致迭代器失效 vector源码3(参考STL源 ...

  10. php -- 4种嵌入标记

    ----- 001-tags.php ----- <!DOCTYPE html> <html> <head> <meta http-equiv="c ...