The Suspects
Time Limit: 1000MS   Memory Limit: 20000K
Total Submissions: 35918   Accepted: 17458

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 代码:
 #include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#define pi acos(-1.0)
#define mj
#define inf 2e8+500
typedef long long ll;
using namespace std;
const int N=3e4+;
int p[N];
int find(int x)
{
return x==p[x]?x:p[x]=find(p[x]);
}
void unit(int x,int y)
{
x=find(x),y=find(y);
if(x==y) return ;
else p[x]=y;
}
int main()
{
int n,m,x,y;
while(scanf("%d%d",&n,&m)==,n||m){
for(int i=;i<n;i++){
p[i]=i;
}
int num;
for(int i=;i<m;i++){
scanf("%d%d",&num,&x);
num--;
while(num--){
scanf("%d",&y);
unit(x,y);
}
}
int ans=;
for(int i=;i<n;i++){
if(find()==find(i)) ans++;
}
printf("%d\n",ans);
}
return ;
}
 

poj 1611 dsu的更多相关文章

  1. poj 1611(并查集)

    http://poj.org/problem?id=1611 题意:有个学生感染病毒了,只要是和这个学生接触过的人都会感染,而和这些被感染者接触的人,也会被感染,现在给定你一些协会的人数,以及所在学生 ...

  2. poj 1611 The Suspects 解题报告

    题目链接:http://poj.org/problem?id=1611 题意:给定n个人和m个群,接下来是m行,每行给出该群内的人数以及这些人所对应的编号.需要统计出跟编号0的人有直接或间接关系的人数 ...

  3. poj 1611 The Suspects(简单并查集)

    题目:http://poj.org/problem?id=1611 0号是病原,求多少人有可能感染 #include<stdio.h> #include<string.h> # ...

  4. 【原创】poj ----- 1611 The Suspects 解题报告

    题目地址: http://poj.org/problem?id=1611 题目内容: The Suspects Time Limit: 1000MS   Memory Limit: 20000K To ...

  5. (并查集)The Suspects --POJ --1611

    链接: http://poj.org/problem?id=1611 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=82830#probl ...

  6. POJ - 1611 The Suspects 【并查集】

    题目链接 http://poj.org/problem?id=1611 题意 给出 n, m 有n个人 编号为 0 - n - 1 有m组人 他们之间是有关系的 编号为 0 的人是 有嫌疑的 然后和 ...

  7. 【裸的并查集】POJ 1611 The Suspects

    http://poj.org/problem?id=1611 [Accepted] #include<iostream> #include<cstdio> #include&l ...

  8. 并查集 (poj 1611 The Suspects)

    原题链接:http://poj.org/problem?id=1611 简单记录下并查集的模板 #include <cstdio> #include <iostream> #i ...

  9. [并查集] POJ 1611 The Suspects

    The Suspects Time Limit: 1000MS   Memory Limit: 20000K Total Submissions: 35206   Accepted: 17097 De ...

随机推荐

  1. better-scroll 遇到的问题 3 (transition-group 相关)

    今天在使用vue动画 transition-group 和 better-scroll 的时候,出现了下拉列表不能滚动的问题. 问题描述: 我写了一个scroll的基础组件,组件接受一个data参数, ...

  2. sharepoint2007就地升级2010系列(四)升级数据库

    上一篇我们完成了系统的升级,今天我们来看一下SQL2005X64是如何升级到SQL2008X64的. 首先,我们先停掉所有sharepoint的服务 其实网上的文档并没有写到这一步,但是我个人觉得,要 ...

  3. 仿真DDR3 Controller IP

    一.Creat a new project,generate a new DDR3 IP,Close Project. 二.打开工程文件下的 X_example_design-->simulat ...

  4. 西门子(SIEMENS)软件安装时需要重启的解决方法,regedit restart

    打开注册表(regedit) 删除注册表项 HKEY_LOCAL_MACHINE\System\CurrentControlSet\Control\Session Manage\PendingFile ...

  5. April 18 2017 Week 16 Tuesday

    Every light has darkness to balance it out. 有光明的地方,必定有黑暗予以平衡. I strive to get a balance between life ...

  6. java 网络流 TCP/UDP

    一.ServerSocket java.lang.Object |-java.net.ServerSocket 有子类SSLServerSocket. 此类实现服务器套接字.服务器套接字等待请求通过网 ...

  7. 如何在windows下运行Linux命令?(转载)

    在windows上可以运行或使用linux下面的命令吗?可以,小编今天就来分享怎么样让Windows支持Linux命令,做这些安装和设置后,就可以非常方便的在windows系统中使用linux下面的命 ...

  8. 检测浏览器中是否有Flash插件

    由于IE和非IE浏览器检测方式不同,所以代码如下 function hasPlugin(name){ debugger; name = name.toLowerCase(); for(var i=0; ...

  9. IOS 音频的 使用说明

    说明 ● 简单来说,音频可以分为2种 ● 音效 • 又称“短音频”,通常在程序中的播放时长为1~2秒 • 在应用程序中起到点缀效果,提升整体用户体验 ● 音乐 • 比如游戏中的“背景音乐”,一般播放时 ...

  10. Uva 11582 巨大的斐波那契数 模运算

    题目链接:https://vjudge.net/contest/156903#problem/A 题意:计算 f(a^b)%n 分析: 1.斐波那契数列是 f(i+2) = f(i+1) + f(i) ...