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Can somebody just brief the approach for “Cobbs and mazes” ?
PROBLEM LINK:CodeChef: Practical coding for everyone
@rahul_g
For every entry and exit points you could run Dijkstra and see whats the shortest path to the other entry/exit points except itself. If the shortest path is less than K then the maze Fails. The maze PASSES if the shortest path considering all the entry/exit points is >= K. @cubefreak777
Got it thanks 
Use Floyd–Warshall algorithm for all-pair shortest path, then just check the min paths for all pairs in A.
Yes, I tried it but wouldn’t that be O(n^3) and n is 10^3 so shouldn’t it give a TLE ?
TLE solution : CodeChef: Practical coding for everyone
I can’t understand how did your N^3 solution passed. I was also using the same approach( because earlier constraints written was N<=100) but I got TLE with O(n^3). Then I commented about it, and they said it’s N<=1000, it was typo.
TLE solution : CodeChef: Practical coding for everyone
If tests were bad then why did the same approach of mine give TLE? 
You can optimize your code by using ints instead of int64, and just use 1 matrix, not 2.
value of N has changed to 1000, how can you apply floyd warshall its compexity is O(N^3) alone?
please correct me if I am wrong but please check the constraints again
It was really a nice contest for me , I did 5 questions in 20 mins approx and was at rank 3 one moment, (very cool for me) but then I tried the Bob question , first I implement it in python it gave me TLE and run time errors then I switched to c++ and surprised to me it gave me runtime error and siegev error. I was bit shocked and left the contest 
Hey, this link is for mu submission to Bob question, could you please help me figuring out where I was wrong why it was giving me run time errors?
Is it fine to run Dijkstra for 100 times with a graph of 1000 nodes?
I think yes. if dijkstra complexity is considered to be v^2, then 100* v^2 should pass. Or you could use priority Queue and reduce the dijkstra time complexity to vlog e
I looked into the solutions…the test cases had instances where number of vertices were indeed 1000, but I dont know how the n^3 solution passed. Although c++ solutions with fast i/o were the ones that mostly passed, not all of them though. Also the question had no multiple test cases. It still unclear to me how the n^3 solution passed, which should not have. Many n^3 solutions in JAVA and python got TLE.
Thank you for pointing that out mid-contest mate! I overlooked it!
If you have done the next question…Bob…please explain it
Yes @first_semester for Bob and the cookies what you were expected to do was a pretty straightforward bfs/dfs from source to destination. Out of all the path you had to count the number of paths that had the smallest length. However, you had to take care of the case where there may be multiple paths crossing a single point on the graph. I think thats all that you had to do!
