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.
代码：
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int f[2010],s[2010];
void init(int n)
{
for(int i=0;i<=n;i++)
f[i]=i;
memset(s,0,sizeof(s));
}
int getf(int u)
{
if(f[u]==u)
return u;
return f[u]=getf(f[u]);
}
bool merge(int u,int v)
{
int t1=getf(u);
int t2=getf(v);
if(t1==t2)
return false;
f[t2]=t1;
return true;
}
int main(void)
{
int t,k=0;
ios::sync_with_stdio(false);
cin>>t;
while(t--)
{
int n, m;
cin>>n>>m;
init(n);
int u,v,t1,t2;
int flag=0;
for(int i=1;i<=m;i++)
{
cin>>u>>v;
if(flag)
continue;
t1=getf(u);
t2=getf(v);
if(t1==t2)
{
flag=1;
continue;
}
//以下操作把同类虫子连接起来
if(s[u]==0)
s[u]=v;
else
merge(s[u],v);//若u已经有对象则把u原来的对象和现在的对象归为同一类
//不用担心u的旧对象和新对象是否是同一性别，因为前面的t1==t2已经确定了u和u现在的对象不可能是同一性别
if(s[v]==0)
s[v]=u;
else
merge(s[v],u);
}
cout<<"Scenario #"<<++k<<":"<<endl;
if(flag)
cout<<"Suspicious bugs found!"<<endl<<endl;
else
cout<<"No suspicious bugs found!"<<endl<<endl;
}
return 0;
}