HDOJ 1829 A Bug's Life 및 쿼드 세트

그리고 수집된 강화 훈련
제목은 다음과 같습니다.
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
TUD Programming Contest 2005, Darmstadt, Germany
 
 
Recommend
linle
만약 A->B 그리고 A가 이전에 나타났었다면, A.opper와 B는 반드시 동성 즉 같은 집합 중일 것이다
다른 주의, 제목은 입력 데이터가 끝난 후에야 결과를 출력하도록 요구한다
/* 구조수조opp는 자체번호로 초기화한다. 만약에 A가 B와 관련이 있다면 먼저 find(A),find(B)의 조작을 하고 뿌리가 같으면 A와 B가 같은 집합, 즉 동성애에 속한다는 것을 설명한다. 그렇지 않으면 아래의 조작을 실행한다. 만약에 A의opp가 그 자체라면 A의opp가 초기화되지 않고 B와 같음을 설명한다. 그렇지 않으면 A의opp와 B를 Unition 조작한다. 이와 유사하게 B의opp가 그 자체라면 B의opp가 초기화되지 않고 A와 같음을 설명한다.그렇지 않으면 B의opp와 A를 Unition 조작*/#include#define N 2100 usingnamespace std;struct node { int father; int opper; }hash[N]; void init()//초기화 {int i;for(i=0;i