A Bug's Life

Time Limit: 10000MS   Memory Limit: 65536K
Total Submissions: 45757   Accepted: 14757

题目链接http://poj.org/problem?id=2492

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.

题意:

给出m对关系,他们是互斥的,问给出的关系是否有矛盾。

题解:

种类并查集,可以用带权并查集解决,这里我就多开了两个集合。

思路可以参见我之前的一篇博客:https://www.cnblogs.com/heyuhhh/p/9980330.html

代码如下:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std; const int N = ;
int t,n,m;
int f[N]; int find(int x){
return f[x]==x ? f[x] :f[x]=find(f[x]);
}
bool same(int x,int y){
return find(x)==find(y);
}
void Union(int x,int y){
int fx=find(x),fy=find(y);
f[fx]=fy;
}
int main(){
scanf("%d",&t);int tot=;
while(t--){
tot++;
bool flag=true;
scanf("%d%d",&n,&m);
for(int i=;i<=N-;i++) f[i]=i;
for(int i=,x,y;i<=m;i++){
scanf("%d%d",&x,&y);
if(!flag) continue ;
int fx=find(x),fy=find(y);
if(!same(x,y) && !same(x+n,y+n)){
Union(x,y+n);Union(x+n,y);
}else flag=false;
}
printf("Scenario #%d:\n",tot);
if(flag) puts("No suspicious bugs found!");
else puts("Suspicious bugs found!");
printf("\n");
}
return ;
}

POJ2492:A Bug's Life(种类并查集)的更多相关文章

  1. POJ2492 A Bug's Life —— 种类并查集

    题目链接:http://poj.org/problem?id=2492 A Bug's Life Time Limit: 10000MS   Memory Limit: 65536K Total Su ...

  2. 【POJ】2492 A bug's life ——种类并查集

    A Bug's Life Time Limit: 10000MS   Memory Limit: 65536K Total Submissions: 28211   Accepted: 9177 De ...

  3. poj2492 A Bug's Life【并查集】

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

  4. 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 ...

  5. 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 ...

  6. hdu1829A Bug's Life(种类并查集)

    传送门 关键在于到根节点的距离,如果两个点到根节点的距离相等,那么他们性别肯定就一样(因为前面如果没有特殊情况,两个点就是一男一女的).一旦遇到性别一样的,就说明找到了可疑的 #include< ...

  7. 【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 ...

  8. A Bug's Life(种类并查集)(也是可以用dfs做)

    http://acm.hdu.edu.cn/showproblem.php?pid=1829   A Bug's Life Time Limit:5000MS     Memory Limit:327 ...

  9. 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 ...

随机推荐

  1. 50条大牛C++编程开发学习建议

    每个从事C++开发的朋友相信都能给后来者一些建议,但是真正为此进行大致总结的很少.本文就给出了网上流传的对C++编程开发学习的50条建议,总结的还是相当不错的,编程学习者(不仅限于C++学习者)如果真 ...

  2. 字典树(Trie)的学习笔记

    按照一本通往下学,学到吐血了... 例题1 字典树模板题吗. 先讲讲字典树: 给出代码(太简单了...)! #include<cstdio> #include<cstring> ...

  3. grep用法小结

    用法 grep [OPTIONS] PATTERN [FILE...] grep [OPTIONS] -e PATTERN ... [FILE...] grep [OPTIONS] -f FILE . ...

  4. 【Leetcode】807. Max Increase to Keep City Skyline

    Description In a 2 dimensional array grid, each value grid[i][j] represents the height of a building ...

  5. [bzoj1359][Baltic2009]Candy

    给定N个数对$(T_i,S_i)$,表示时刻$S_i$时在位置$T_i$处出现一粒糖果.有一些机器人可供使用,每个机器人可花费一单位时间向相邻位置移动.要求用最少的机器人接到全部糖果.时刻0时机器人位 ...

  6. 2、Java并发编程:如何创建线程

    Java并发编程:如何创建线程? 在前面一篇文章中已经讲述了在进程和线程的由来,今天就来讲一下在Java中如何创建线程,让线程去执行一个子任务.下面先讲述一下Java中的应用程序和进程相关的概念知识, ...

  7. 云计算之路-阿里云上:受够了OCS,改用ECS+Couchbase跑缓存

    当今天早上在日志中发现这样的错误之后,对阿里云OCS(mecached缓存服务)的积怨倾泻而出. 2014-06-08 07:15:56,078 [ERROR] Enyim.Caching.Memca ...

  8. Java Set集合(HashSet、TreeSet)

    什么是HashSet?操作过程是怎么样的? 1.HashSet底层实际上是一个HashMap,HashMap底层采用了哈希表数据结构 2.哈希表又叫做散列表,哈希表底层是一个数组,这个数组中每一个元素 ...

  9. Linux-Shell脚本编程-学习-5-Shell编程-使用结构化命令-if-then-else-elif

    if-then语句 if-then语句格式如下 if comman then command fi bash shell中的if语句可鞥会和我们接触的其他if语句的工作方式不同,bash shell的 ...

  10. C++学习002-C++代码中插入汇编语句

    在C++中我们有时会遇到使用汇编语言的情况,这时可以在前面加上关键字“_asm”宏. 如下示例 编写环境 :vs2015 int main() { __asm mov al, 0x20; __asm ...