POJ 2492 A Bug's Life (并查集)-白红宇

POJ 2492 A Bug's Life (并查集)

阅读量：6910 次

发布时间：2019-06-27

本文共 3296 字，大约阅读时间需要 10 分钟。

A Bug's Life

Time Limit: 10000MS		Memory Limit: 65536K
Total Submissions: 30130		Accepted: 9869

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

23 31 22 31 34 21 23 4

Sample Output

Scenario #1:Suspicious bugs found!Scenario #2:No suspicious bugs found! 判断两只虫子是否同性。

1 #include 
     
        2 #include 
      
         3 #include 
       
          4 #include 
        
           5 #include 
         
            6 #include 
           7 #include 
           
             8 #include 
            
              9 #include 
             
               10 #include 
              
                11 #include 
               
                 12 #include 
                
                  13 using namespace std; 14 15 const int SIZE = 2005; 16 int FATHER[SIZE],MARK[SIZE],RANK[SIZE]; 17 18 void ini(int); 19 int get_father(int); 20 void unite(int,int); 21 bool same(int,int); 22 23 int main(void) 24 { 25 int t,n,m,x,y,count = 0; 26 bool flag; 27 28 scanf("%d",&t); 29 while(t --) 30 { 31 count ++; 32 scanf("%d%d",&n,&m); 33 ini(n); 34 flag = false; 35 while(m --) 36 { 37 scanf("%d%d",&x,&y); 38 if(flag) 39 continue; 40 if(same(x,y)) 41 flag = true; 42 if(!MARK[x] && !MARK[y]) 43 { 44 MARK[x] = y; 45 MARK[y] = x; 46 } 47 else if(!MARK[x]) 48 { 49 MARK[x] = y; 50 unite(x,MARK[y]); 51 } 52 else if(!MARK[y]) 53 { 54 MARK[y] = x; 55 unite(y,MARK[x]); 56 } 57 else 58 { 59 unite(x,MARK[y]); 60 unite(y,MARK[x]); 61 } 62 } 63 printf("Scenario #%d:\n",count); 64 if(flag) 65 puts("Suspicious bugs found!"); 66 else 67 puts("No suspicious bugs found!"); 68 puts(""); 69 } 70 71 72 return 0; 73 } 74 75 void ini(int n) 76 { 77 for(int i = 1;i <= n;i ++) 78 { 79 MARK[i] = RANK[i] = 0; 80 FATHER[i] = i; 81 } 82 } 83 84 int get_father(int n) 85 { 86 if(n == FATHER[n]) 87 return n; 88 return FATHER[n] = get_father(FATHER[n]); 89 } 90 91 void unite(int x,int y) 92 { 93 x = get_father(x); 94 y = get_father(y); 95 96 if(x == y) 97 return ; 98 if(RANK[x] < RANK[y]) 99 FATHER[x] = y;100 else101 {102 FATHER[y] = x;103 if(RANK[x] == RANK[y])104 RANK[x] ++;105 }106 }107 108 bool same(int x,int y)109 {110 return get_father(x) == get_father(y);111 }

转载于:https://www.cnblogs.com/xz816111/p/4536825.html

你可能感兴趣的文章