CAM5  prayatna PR
help the Prayatna pr team
Well, the annual technical symposium of Department of Computer Technology is around the corner. All that we need, to make it a grand success is Publicity among the peer groups ( ofCourse the sponsors too :P ). We decided to share the job between the student groups. As per the plan we decided to meet people in person and influence them to attend Prayatna. But to meet them we have to go to various student groups. To do so, we had to cut our classes. But being studious. students refused to cut more classes. Instead of meeting every one in person we decided to meet few people such that the person to whom we pass the news will spread it to all his friends. And those friends will pass it to other friends and so on. Your task is to find the number of people to be met by the organizers to spread the news.
Caution: Large I/O
Input
First line of input is 't'  Test cases. Follwed by N, the number of peers in the testcase ( 0 to N1 ). followed by the number of friend description 'e'. Following are 'e' descriptions of type "a b" denoting 'a' friends with 'b'. If 'a' is friends with 'b' then 'b' is friends with 'a'.
Output
Output contains t line, the number of people, the organizers have to meet in person for each test case.
Example
Input: 2 4 2 0 1 1 2 3 0 Output: 2 3
Explanation
case 1 : 0 is friends with 1; 1 is friends with 2; so if we pass the news to 0 & 3, news will pass it to the entire N peers.
case 2 : no one is friends with any one. So we have to meet every one in person.
Constraints
t = 10
2 <= N <= 100000
0 <= e <= N/2
hide comments
saurabh_kl:
20210202 07:28:36
Don't consider line break for testcases(i.e don' t use nextLine() in Java for a blank line (otherwise you will get runtime error) 

wille_25:
20201230 12:22:46
Last edit: 20210315 13:44:12 

baukali22:
20200811 12:13:46
Don't forget to clear the adjacency list after each testcase


noobiesag_0206:
20200718 01:44:06
@rks0023 don't cry my son 

rks0023:
20200708 22:26:47
:( Last edit: 20200708 22:27:57 

kumar_anubhav:
20200630 19:41:31
Ac in one Go!!


picchi_67:
20200613 20:00:44
basic dsu problem...count the number of different parent. 

suyasha_:
20200509 11:12:24
simple dfs, count the number of connected components. 

manish_thakur:
20200418 20:29:13
simple dfs


sonalisingh:
20200316 09:18:23
thanks @deependra_18 your comment was helpful :) 
Added by:  karthikeyan 
Date:  20120111 
Time limit:  1s 
Source limit:  50000B 
Memory limit:  1536MB 
Cluster:  Cube (Intel G860) 
Languages:  All except: ASM64 
Resource:  own problem 