http://poj.org/problem?id=1611

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

0号患了病毒,只要和0号一组的人同样会患病毒。。。。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int bin[];
int q[];
int n,m;
int cmp(const void *a,const void *b)
{
return *(int *)a-*(int *)b;
}
int findx(int x)
{
int r,j,k;
r=x;
while(bin[r]!=r)
r=bin[r];
k=x;
while(k!=r)
{
j=bin[k];
bin[k]=r;
k=j; }
return r;
}
void merge(int x,int y)
{
int fx,fy;
fx=findx(x);
fy=findx(y);
if(fx!=fy)
{
bin[fx]=fy;
}
} int main()
{
int n,x,t,i,j;
while(scanf("%d%d",&n,&x)!=EOF)
{
if(n==&&x==) break;
for(i=; i<n; i++)
bin[i]=i;
while(x--)
{
scanf("%d",&t);
memset(q,,sizeof(q));
for(j=; j<t; j++)
scanf("%d",&q[j]);
qsort(q,t,sizeof(q[]),cmp);
for(j=; j<t; j++)
{
merge(q[j],q[j-]);
} }
/*for(i=0;i<n;i++)
printf("...%d",bin[i]);*/
int l=;
for(i=; i<n; i++)
{
if(findx(i)==findx())
l++; }
printf("%d\n",l); }
return ;
}

POJ1611:The Suspects(模板题)的更多相关文章

  1. POJ1611 The Suspects 并查集模板题

    题目大意:中文题不多说了 题目思路:将每一个可能患病的人纳入同一个集合,然后遍历查找每个点,如果改点点的根节点和0号学生的根节点相同,则该点可能是病人. 模板题并没有思路上的困难,只不过在遍历时需要额 ...

  2. [AHOI 2009] 维护序列(线段树模板题)

    1798: [Ahoi2009]Seq 维护序列seq Time Limit: 30 Sec  Memory Limit: 64 MB Description 老师交给小可可一个维护数列的任务,现在小 ...

  3. HDU 2222 AC自动机模板题

    题目: http://acm.hdu.edu.cn/showproblem.php?pid=2222 AC自动机模板题 我现在对AC自动机的理解还一般,就贴一下我参考学习的两篇博客的链接: http: ...

  4. POJ2774 & 后缀数组模板题

    题意: 求两个字符串的LCP SOL: 模板题.连一起搞一搞就好了...主要是记录一下做(sha)题(bi)过程心(cao)得(dan)体(xin)会(qing) 后缀数组概念...还算是简单的,过程 ...

  5. HDU 1251 Trie树模板题

    1.HDU 1251 统计难题  Trie树模板题,或者map 2.总结:用C++过了,G++就爆内存.. 题意:查找给定前缀的单词数量. #include<iostream> #incl ...

  6. HDU-3549 最大流模板题

    1.HDU-3549   Flow Problem 2.链接:http://acm.hdu.edu.cn/showproblem.php?pid=3549 3.总结:模板题,参考了 http://ww ...

  7. HDU 4280:Island Transport(ISAP模板题)

    http://acm.hdu.edu.cn/showproblem.php?pid=4280 题意:在最西边的点走到最东边的点最大容量. 思路:ISAP模板题,Dinic过不了. #include & ...

  8. HDU-2222 Keywords Search(AC自动机--模板题)

    题目大意:统计一共出现了多少次模板串. 题目分析:AC自动机的模板题.不过这题有坑,相同的模板串不能只算一次. 代码如下: # include<iostream> # include< ...

  9. Dancing Link --- 模板题 HUST 1017 - Exact cover

    1017 - Exact cover Problem's Link:   http://acm.hust.edu.cn/problem/show/1017 Mean: 给定一个由0-1组成的矩阵,是否 ...

  10. AC自动机 - 多模式串匹配问题的基本运用 + 模板题 --- HDU 2222

    Keywords Search Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)T ...

随机推荐

  1. DDoS(Distributed Denial of Service,分布式拒绝服务)

    DDoS:Distributed Denial of Service,即分布式拒绝服务攻击. 借助于客户/服务器技术,将多个计算机联合起来作为攻击平台,对一个或多个目标发动DDoS攻击,从而成倍地提高 ...

  2. nodejs yarn包管理工具

    Yarn https://yarnpkg.com/zh-Hans/docs/install#windows-stable 安装包 Yarn Npm yarn npm i yarn global add ...

  3. 在Ubuntu上安装Chrome Driver和Firefox Driver

    在Ubuntu上安装Chrome Driver和Firefox Driver 此文章只介绍Chrome Driver(Firefox Driver和该步骤相同) 下载链接:http://chromed ...

  4. python列表操作方法

    系统的列表操作方法不加赘述,这里增添一些列表操作技巧: 1.利用sum函数把多元列表变成一元: >>> texts_filtered_stopwords [['writing', ' ...

  5. Win10下安装MySQL5.6

    Win10下安装MySQL5.6 我分了两种下载安装的方式给大家看,注意数据库这个东西不在乎版本是不是最新,在乎的是够稳定,现在公司中常用的是mysql5.5和mysql5.6的版本,我现在就用mys ...

  6. python数据结构之图论

    本篇学习笔记内容为图的各项性质.图的表示方法.图ADT的python实现 图(Graph) 是数据结构和算法学中最强大的框架之一(或许没有之一).图几乎可以用来表现所有类型的结构或系统,从交通网络到通 ...

  7. cinder 和 qt5 vs2015结合

    下载编译好的cinder_0.9.1_vc2013, 用vs2015打开 cinder_0.9.1_vc2013\proj\vc2013\cinder.sln 重新编译 由于我用的qt也是vs2015 ...

  8. SHOW PROCESSLIST shows which threads are running 查看线程 解决瓶颈

    小结: 1.查看全部线程: https://dev.mysql.com/doc/refman/8.0/en/show-processlist.html MySQL 8.0 Reference Manu ...

  9. iOS-静态库,动态库,framework浅析(二)

    创建.a静态库 第一步,新建工程.     一般使用工程名就使用库的名称,比如我这里用FMDB来创建静态库,我的工程名就取名为FMDB,创建的.a静态库就是libFMDB.a.             ...

  10. php性能提升与检测

    1.使用xhprof分析器检测性能各种消耗 2.php-fpm中进程池的配置参数查看最大进程数.进程最大处理http请求量.进程时间过多的http请求.每个进程使用的最大内存. 参考地址:https: ...