A Bug's Life

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9223    Accepted Submission(s): 2965

Problem Description
Background 
Professor Hopper is researching the sexual behavior of a rare species of bugs. He assumes that they feature two different genders and that they only interact with bugs of the opposite gender. In his experiment, individual bugs and their interactions were easy to identify, because numbers were printed on their backs.

Problem 
Given a list of bug interactions, decide whether the experiment supports his assumption of two genders with no homosexual bugs or if it contains some bug interactions that falsify it.

 
Input
The first line of the input contains the number of scenarios. Each scenario starts with one line giving the number of bugs (at least one, and up to 2000) and the number of interactions (up to 1000000) separated by a single space. In the following lines, each interaction is given in the form of two distinct bug numbers separated by a single space. Bugs are numbered consecutively starting from one.
 
Output
The output for every scenario is a line containing "Scenario #i:", where i is the number of the scenario starting at 1, followed by one line saying either "No suspicious bugs found!" if the experiment is consistent with his assumption about the bugs' sexual behavior, or "Suspicious bugs found!" if Professor Hopper's assumption is definitely wrong.
 
Sample Input
2
3 3
1 2
2 3
1 3
4 2
1 2
3 4
 
Sample Output
 
Scenario #1:
Suspicious bugs found!
 
Scenario#2:
No suspicious bugs found!
 
Hint

Huge input,scanf is recommended.

Source
 #include<stdio.h>
#include<string.h>
int f[];
int dis[]; int find(int x)
{
int t;
if(f[x]==x)
return x; t=find(f[x]);
dis[x]=(dis[x]+dis[f[x]])%;
f[x]=t;
return t; } int make(int x,int y)
{
int a=find(x);
int b=find(y);
if(a==b)
{
if(dis[x]==dis[y])
return ;
return ;
} f[a]=b;
dis[a]=(dis[x]+dis[y]+)%;
return ; } int main()
{
//freopen("in.txt","r",stdin);
int t,count=;
int n,m;
int i,a,b,flag;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
memset(dis,,sizeof(dis)); for(i=;i<=n;i++)
f[i]=i; flag=;
for(i=;i<m;i++)
{
scanf("%d%d",&a,&b);
if(make(a,b))
{
flag=;
}
} printf("Scenario #%d:\n",count++);
if(flag)
printf("Suspicious bugs found!\n");
else
printf("No suspicious bugs found!\n");
printf("\n");
}
return ;
}
 

1829 A Bug's Life的更多相关文章

  1. HDU 1829 A Bug's Life (种类并查集)

    传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1829 A Bug's Life Time Limit: 15000/5000 MS (Java/Oth ...

  2. hdu 1829 A Bug's Life(分组并查集(偏移量))

    A Bug's Life Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Tot ...

  3. hdu - 1829 A Bug's Life (并查集)&&poj - 2492 A Bug's Life && poj 1703 Find them, Catch them

    http://acm.hdu.edu.cn/showproblem.php?pid=1829 http://poj.org/problem?id=2492 臭虫有两种性别,并且只有异性相吸,给定n条臭 ...

  4. 【进阶——种类并查集】hdu 1829 A Bug's Life (基础种类并查集)TUD Programming Contest 2005, Darmstadt, Germany

    先说说种类并查集吧. 种类并查集是并查集的一种.但是,种类并查集中的数据是分若干类的.具体属于哪一类,有多少类,都要视具体情况而定.当然属于哪一类,要再开一个数组来储存.所以,种类并查集一般有两个数组 ...

  5. hdu 1829 A Bug's Life(并查集)

                                                                                                    A Bu ...

  6. HDU 1829 - A Bug's Life

    Problem Description Background Professor Hopper is researching the sexual behavior of a rare species ...

  7. HDU 1829 A Bug's Life 【带权并查集/补集法/向量法】

    Background Professor Hopper is researching the sexual behavior of a rare species of bugs. He assumes ...

  8. hdoj 1829 A bug's life 种类并查集

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1829 并查集的一个应用,就是检测是否存在矛盾,就是两个不该相交的集合有了交集.本题就是这样,一种虫子有 ...

  9. HDU 1829 A Bug's Life(种类并查集)

    思路:见代码吧. #include <stdio.h> #include <string.h> #include <set> #include <vector ...

随机推荐

  1. 创建本地Ubuntu镜像

    参考文档 http://www.howtoforge.com/local_debian_ubuntu_mirror 安装服务 : sudo apt-get install apt-mirror apa ...

  2. Android模拟器中安装APK文件(转)

    1.平台环境:Win7系统, 安装Eclipse,android4.0(sdk) 2.随便创建个工程(HelloWorld),结果如下: 3.运行(Run HelloWorld),启动模拟器,如下所示 ...

  3. 取代file_get_contents 的一个采集函数

    function url_get_content($url=''){ $ch = curl_init(); $timeout = 100; $browser = 'Mozilla/5.0 (Windo ...

  4. [转]Linux 分区 swap

    如何合理设置Linux的swap分区 原创作品,允许转载,转载时请务必以超链接形式标明文章 原始出处 .作者信息和本声明.否则将追究法律责任.http://commandos.blog.51cto.c ...

  5. Hadoop上路-01_Hadoop2.3.0的分布式集群搭建

    一.配置虚拟机软件 下载地址:https://www.virtualbox.org/wiki/downloads 1.虚拟机软件设定 1)进入全集设定 2)常规设定 2.Linux安装配置 1)名称类 ...

  6. Unity3d 动态批处理的问题

    这段时间做unity3d的优化,主要的入手是减少draw call.    1.代码上主要是把一些零碎的同材质的合并成一个大的mesh.    2.减少不必要的全屏后期处理.把摄像机的renderin ...

  7. ASP.NET中的ViewState

    曾经在两次面试中都遇到了这个问题,就是ViewState中存储的变量到底存储在哪里.由于基础比较差,以前在学习的时候,就没有注意 到这里的细节,包括Session中存储的变量,所以我想ViewStat ...

  8. Oracle中的IF...THEN...ELSE判断

    if...then...else是最常见的一种判断语句,他可以实现判断两种情况. 标准语法如下: if <condition_expression> then plsql_sentence ...

  9. oracle-11g-配置dataguard

    1.环境信息:系统:oracle-linux 5.7 数据库版本:Oracle Database 11g Enterprise Edition Release 11.2.0.3.0 - 64bit P ...

  10. iOS 进阶 第九天(0408)

    0408 makekeyAndVisible解释 一个程序可以有多个Window,但只有一个窗口能够成为主窗口.如图中所示,此时的window2是主窗口.主窗口用处大了.从iOS7开始无论是主窗口还是 ...