BestCoder Round #48 ($)HDU5285 wyh2000 and pupil

wyh2000 and pupil

                                                    Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
                                                                                Total Submission(s): 1239    Accepted Submission(s): 384

Problem Description

Young theoretical computer scientist wyh2000 is teaching his pupils.

Wyh2000 has n pupils.Id of them are from  1 to  n.In order to increase the cohesion between pupils,wyh2000 decide to divide them into 2 groups.Each group has at least 1 pupil.

Now that some pupils don't know each other(if  a doesn't know  b,then  b doesn't know  a).Wyh2000 hopes that if two pupils are in the same group,then they know each other,and the pupils of the first group must be as much as possible.

Please help wyh2000 determine the pupils of first group and second group. If there is no solution, print "Poor wyh".

 

Input

In the first line, there is an integer  T indicates the number of test cases.

For each case, the first line contains two integers  n,m  indicate the number of pupil and the number of pupils don't konw each other.

In the next m lines,each line contains 2 intergers  x,y(x < y) ,indicates that  x don't know  y and  y don't know  x,the pair  (x,y)  will only appear once.

T≤10,0≤n,m≤100000

 

Output

For each case, output the answer.

 

Sample Input

  

   2
8 5
3 4
5 6
1 2
5 8
3 5
5 4
2 3
4 5
3 4
2 4
  

 

Sample Output

  

   5 3
Poor wyh
  

 

Source

BestCoder Round #48 ($)

出题人：

如果$a$不认识$b$，那么在$a,b$间连一条边，这样有解当且仅当这张图是二分图。

由于可能有多个二分图，而题目要求第一组的人尽可能多，所以贪心的选择即可。

要注意$m=0$的情况。

#include 
  
   
#include 
   
    
#include 
    
     
#include 
     
      
#include 
      
       
#include 
       
         #include