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.
 
 
 
 
数据量大要用scanf

#include<iostream>
#include<map>
#include<string>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<cstring>
#include<cmath>
#include<vector>
#include<set>
#include<queue>
#include<iomanip>
#include<iostream>
using namespace std;
#define MAXN 2003
#define INF 0x3f3f3f3f
typedef long long LL;
/*带权并查集,
虫子只和不同性别交配,所以并查集距离为1,如果有不匹配,就标记
*/
int pre[MAXN],rank[MAXN],n,m;
int find(int x)
{
if(pre[x]==-)
return x;
int fx = pre[x];
pre[x] = find(pre[x]);
rank[x] = (rank[x]+rank[fx])%;
return pre[x];
}
bool mix(int x,int y)
{
int fx = find(x),fy=find(y);
if(fx==fy)
{
if(rank[x]!=(rank[y]+)%)
return false;
return true;
}
pre[fy] = fx;
rank[fy] = (rank[x]-rank[y]++)%;
return true;
}
void Init()
{
memset(rank,,sizeof(rank));
memset(pre,-,sizeof(pre));
}
int main()
{
int t;
scanf("%d",&t);
for(int i=;i<=t;i++)
{
scanf("%d%d",&n,&m);
Init();
int x,y;
bool f = false;
for(int j=;j<m;j++)
{
scanf("%d%d",&x,&y);
if(f) continue;
if(!mix(x,y))
f = true;
}
printf("Scenario #%d:\n",i);
if(f)
printf("Suspicious bugs found!\n");
else
printf("No suspicious bugs found!\n");
printf("\n");
}
return ;
}

J - A Bug's Life 并查集的更多相关文章

  1. nyoj 209 + poj 2492 A Bug's Life (并查集)

    A Bug's Life 时间限制:1000 ms  |  内存限制:65535 KB 难度:4   描述 Background  Professor Hopper is researching th ...

  2. 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 ...

  3. hdu 1829 A Bug's Life(并查集)

                                                                                                    A Bu ...

  4. poj 2492A Bug's Life(并查集)

    /* 目大意:输入一个数t,表示测试组数.然后每组第一行两个数字n,m,n表示有n只昆虫,编号从1—n,m表示下面要输入m行交配情况,每行两个整数,表示这两个编号的昆虫为异性,要交配. 要求统计交配过 ...

  5. POJ 2492 A Bug's Life 并查集的应用

    题意:有n只虫子,每次给出一对互为异性的虫子的编号,输出是否存在冲突. 思路:用并查集,每次输入一对虫子后就先判定一下.如果两者父亲相同,则说明关系已确定,再看性别是否相同,如果相同则有冲突.否则就将 ...

  6. 牛客网多校第4场 J Hash Function 【思维+并查集建边】

    题目链接:戳这里 学习博客:戳这里 题意: 有n个空位,给一个数x,如果x%n位数空的,就把x放上去,如果不是空的,就看(x+1)%n是不是空的. 现在给一个已经放过数的状态,求放数字的顺序.(要求字 ...

  7. poj2492_A Bug's Life_并查集

    A Bug's Life Time Limit: 10000MS   Memory Limit: 65536K Total Submissions: 34947   Accepted: 11459 D ...

  8. [poj2492]A Bug's Life(并查集+补集)

    A Bug's Life Time Limit: 10000MS   Memory Limit: 65536K Total Submissions: 34678   Accepted: 11339 D ...

  9. POJ 2492 A Bug's Life (并查集)

    A Bug's Life Time Limit: 10000MS   Memory Limit: 65536K Total Submissions: 30130   Accepted: 9869 De ...

随机推荐

  1. http升级https(转)

    让你的网站免费支持 HTTPS 及 Nginx 平滑升级 为什么要使用 HTTPS ? 首先来说一下 HTTP 与 HTTPS 协议的区别吧,他们的根本区别就是 HTTPS 在 HTTP 协议的基础上 ...

  2. loadrunner11 安装破解,汉化包

    说说自己的心痛史,好不容易安装了loadrunner 11 居然浏览器不支持,我的系统是win8.1,ie浏览器最低支持ie11,我还能说啥子...其他浏览器试过了依旧是不可以!!所以我安装了一个虚拟 ...

  3. SVN配置详解

    原文:http://swjr.blog.com.cn/archives/2006/TheRoadToSubversion1authz.shtml http://www.dayuer.com/freeb ...

  4. ASP.NET SQL 总结

    1.SQLSERVER服务器中,给定表 table1 中有两个字段 ID.LastUpdateDate,ID表示更新的事务号, LastUpdateDate表示更新时的服务器时间,请使用一句SQL语句 ...

  5. FFT学习及简单应用(一点点详细)

    什么是FFT 既然打开了这篇博客,大家肯定都已经对FFT(Fast Fourier Transformation)有一点点了解了吧 FFT即为快速傅里叶变换,可以快速求卷积(当然不止这一些应用,但是我 ...

  6. N - Binomial Showdown (组合数学)

    Description In how many ways can you choose k elements out of n elements, not taking order into acco ...

  7. ACM_校庆素数

    校庆素数 Time Limit: 2000/1000ms (Java/Others) Problem Description: 广财建校33年了,如今迎来了她的校庆. 小财最近想在研究素数,她突发奇想 ...

  8. 【SpringMVC框架】非注解的处理器映射器和适配器

    参考来源:     http://blog.csdn.net/acmman/article/details/46968939 处理器映射器就是根据URL来找Handler,处理器适配器就是按照它要求的 ...

  9. phpstudy初级总结

    1.问题一 问题症状:访问http://localhost/phpMyWind/install/不出现安装或登录页面 考虑一下情况: 1.是否打开了PHPstudy, (当Apache不能启用时,考虑 ...

  10. android视频播放器系列(二)——VideoView

    最近在学习视频相关的知识,现在也是在按部就班的一步步的来,如果有同样需求的同学可以跟着大家一起促进学习. 上一节说到了可以使用系统播放器以及浏览器播放本地以及网络视频,但是这在很大程度上并不能满足我们 ...