A Bug's Life

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11981    Accepted Submission(s): 3901

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.

 
并查集题目。
记录一下当前结点到根结点的距离。 当再次找进行”并“操作时,判断属于同一根节点的两个节点到根节点的距离,同时为奇数或偶数,那么久存在矛盾。
  

/* ***********************************************
Author :pk28
Created Time :2015/8/15 9:54:22
File Name :4.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 3000+10
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << ;
const double eps=1e-;
using namespace std; bool cmp(int a,int b){
return a>b;
}
int fa[maxn];
int d[maxn];
int n,m,mark;
void init(){
for(int i=;i<=n+;i++){
d[i]=;
fa[i]=i;
}
mark=;
}
int findfa(int x){
if(x==fa[x])return x;
else{
int root=findfa(fa[x]);
d[x]+=d[fa[x]];//记录到根节点的距离
fa[x]=root;
return fa[x];
}
}
void Union(int a,int b){
int x=findfa(a);
int y=findfa(b);
if(x==y){
if((d[a]&)==(d[b]&))mark=;
return ;
}
else{
fa[x]=y;
d[x]=(1+d[b]-d[a]);//向量
}
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
//freopen("out.txt","w",stdout);
int T;
cin>>T;
int a,b;
int cnt=;
while(T--){
scanf("%d%d",&n,&m);
init();
for(int i=;i<m;i++){
scanf("%d %d",&a,&b);
if(mark)continue;
Union(a,b);
}
printf("Scenario #%d:\n",cnt++);
if(mark)printf("Suspicious bugs found!\n");
else printf("No suspicious bugs found!\n");
printf("\n");
}
return ;
}

HDU 1829/POJ 2492 A Bug's Life的更多相关文章

  1. 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 ...

  2. hdu 1829 &amp;poj 2492 A Bug&#39;s Life(推断二分图、带权并查集)

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

  3. POJ 2492 A Bug's Life(并查集)

    http://poj.org/problem?id=2492 题意 :就是给你n条虫子,m对关系,每一对关系的双方都是异性的,让你找出有没有是同性恋的. 思路 :这个题跟POJ1703其实差不多,也是 ...

  4. (简单) POJ 2492 A Bug's Life,二分染色。

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

  5. POJ 2492 A Bug's Life(带权并查集)

    题目链接:http://poj.org/problem?id=2492 题目大意:有n只虫子,m对关系,m行每行有x y两个编号的虫子,告诉你每对x和y都为异性,先说的是对的,如果后面给出关系与前面的 ...

  6. POJ 2492 A Bug's Life

    传送门:A Bug's Life Description Background Professor Hopper is researching the sexual behavior of a rar ...

  7. 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 ...

  8. POJ 2492 A Bug's Life (并查集)

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

  9. POJ 2492 A Bug's Life【并查集高级应用+类似食物链】

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

随机推荐

  1. js中cookie、sessionStorage、localStorage

    一.cookie <!DOCTYPE html> <html> <head> <meta charset="utf-8"> < ...

  2. (6)C#事务处理

    为了方便移到了ADO.NET分类里 事务的主要特征是,任务要么全部完成,要么都不完成 事务常用于写入或更新数据库中的数据.将数据写入文件或注册表也可以使用事物. ADO.NET不支持跨越多个连接的事物 ...

  3. [原创][SW]一些实用软件的小tips(长期更新)

    0. 简介 生活中我们经常使用许多的小工具或软件,来提高我们的工作效率,比如UltraEdit.Notepad++等.本文主要做一些记录,目的呢就是防止自己遗忘或者是快速的查询,来源是自己的摸索和网络 ...

  4. ubuntu系统克隆

    使用clonezilla,原文地址:http://www.linuxidc.com/Linux/2014-09/107117.htm 类似的一篇:http://storysky.blog.51cto. ...

  5. 某考试 T3 sine

    推完一波式子之后发现是个矩阵23333. 其实只要发现是矩阵之后就是个水题了. #include<bits/stdc++.h> #define ll long long using nam ...

  6. PAT甲级练习题1001、1002

    1001 A+B Format (20 分)   Calculate a+b and output the sum in standard format -- that is, the digits ...

  7. OpenGL - Tessellation Shader 【转】

    http://blog.sina.com.cn/s/blog_8c7d49f20102v4qm.html Patch is just an ordered list of vertices (在tes ...

  8. AngularJS的简单表单验证

    代码下载:https://files.cnblogs.com/files/xiandedanteng/angularjsCheckSimpleForm.rar 代码: <!DOCTYPE HTM ...

  9. 前端学习——使用Ajax方式POST JSON数据包

    0.前言     本文解释怎样使用Jquery中的ajax方法传递JSON数据包,传递的方法使用POST(当然PUT又有时也是一个不错的选择).POST JSON数据包相比标准的POST格式可读性更好 ...

  10. icmp的程序(ping的实现)

    code来源于<网络编程与分层协议设计> chap7 ICMP协议程序设计 ----没有理解,没有编译,只是敲了出来 ping.h #define ICMP_ECHOREPLY 0#def ...