A Bug‘s Life

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8224    Accepted Submission(s): 2631

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!

这道题很像算法艺术与分析上的并查集那一部分的第二题。。额。。好像是这道题。。

假设a 和b 是异性，b和c是异性， 那么a和c就是同性，所以所有的虫子分成两部分即a的同性，和a的异性。

那么可以用数组存储和a同性的，数组为Q[]，那么就有Q[a]=b。若又出现a和d是异性，那么就把Q[a]和d并在一棵树里即unsion(Q[a],d)，因为是相互的，所以也有unsion(Q[d],a);

最终就成了两个集合，若输入a b在同一集合中就意味着a和b是同性恋。。。

代码：

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <algorithm>
 4 #include <queue>
 5 #include <iostream>
 6 using namespace std;
 7
 8 int father[1000005];
 9 int Q[1000005];
10
11 int findroot(int p)
12 {
13     int r=p;
14     while(r!=father[r])
15     r=father[r];
16     return r;
17 }
18
19 void unsion(int p,int q)
20 {
21     int fp=findroot(p);
22     int fq=findroot(q);
23     father[fq]=fp;
24 }
25 main()
26 {
27     int n, m, i, j, k=1;
28     int t;
29     cin>>t;
30     while(t--)
31     {
32         scanf("%d %d",&n,&m);
33         int flag=0, x, y;
34         for(i=1;i<=n;i++) father[i]=i;
35         memset(Q,0,sizeof(Q));
36         while(m--)
37         {
38             scanf("%d %d",&x,&y);
39             if(flag) continue;
40             int fx=findroot(x);
41             int fy=findroot(y);
42             if(fx==fy)
43             {
44                 flag=1;
45                 continue;
46             }
47             if(!Q[x]) Q[x]=y;
48             else unsion(Q[x],y);
49             if(!Q[y]) Q[y]=x;
50             else unsion(Q[y],x);
51         }
52         printf("Scenario #%d:\n",k++);
53         if(flag) cout<<"Suspicious bugs found!"<<endl<<endl;
54         else cout<<"No suspicious bugs found!"<<endl<<endl;
55     }
56 }

HDU 1829 并查集