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. 重复造轮子之RSA算法(一) 大素数生成

    出于无聊, 打算从头实现一遍RSA算法 第一步, 大素数生成 Java的BigInteger里, 有个现成的方法 public static BigInteger probablePrime(int ...

  2. Codeforces 946 A.Partition

    随便写写,然后写D的题解. A. Partition   time limit per test 1 second memory limit per test 256 megabytes input ...

  3. fread函数和fwrite函数

    1.函数功能   用来读写一个数据块. 2.一般调用形式   fread(buffer,size,count,fp);   fwrite(buffer,size,count,fp); 3.说明   ( ...

  4. L1-1. 出生年【STL放的位置】

    L1-1. 出生年 时间限制 400 ms 内存限制 65536 kB 代码长度限制 8000 B 判题程序 Standard 作者 陈越 以上是新浪微博中一奇葩贴:“我出生于1988年,直到25岁才 ...

  5. [Inside HotSpot] UseParallelGC和UseParallelOldGC的区别

    JVM的很多参数命名很有迷惑性,-XX:+UseParallel,-XX:+UseParallelOldGC,-XX:+UseParNewGC,-XX:+UseConcMarkSweepGC咋一看容易 ...

  6. 在路上:安全公司“跨界”SD-WAN

    编者按:本文是SDNLAB“企业+”特别报道之一.“企业+”是SDNLAB重点打造的栏目,汇聚信息行业运营商.设备商.互联网公司.软件公司.集成公司.融创投资公司.科研院所等企业,重新定义IT行业撮合 ...

  7. javascript 函数初探 (三)--- javascript 变量的作用域

    javascript 变量的作用域: 这是一个至关重要的问题.特别是当我们从别的语言转向javascript时,必须要明白一点,即在javascript中,变量的定义并不是以代码块作为作用域的,而是以 ...

  8. html中的列表标签

    1.<dl>定义列表,<dt>定义列表中的项目,<dd>对项目的描述 例: 效果: 2.<ul>无序列表,<li>列表项 例: 效果: 3. ...

  9. POCO类

    我认为POCO(简单传统CLR对象)重点应该是简单,不跟其他不相关的类进行关联关系或不相关的属性.<NHibernate 4 Beginner Guid>这本书有提到,应该是满足下面三个条 ...

  10. ARM Holdings

    http://en.wikipedia.org/wiki/Advanced_RISC_Machines ARM Holdings  (Redirected from Advanced RISC Mac ...