himanshujaju's blog

By himanshujaju, history, 3 years ago, In English,

On behalf of Topcoder India and hmehta, allow me to welcome you to Topcoder Collegiate India Contest, 2017.

The Qualification Round will be held on 29th July, 2017 at 09:00 pm IST. Top 10 from the Online Round Will Qualify for the Final Round which will take place during the TCO17 India Regionals event. Note that you must be a student studying in an Indian university to be eligible for the next round. There is a fun round in parallel to this event which is open for all participants. Both these rounds are rated!

The Final Onsite Round will be held on 20th August, 2017 at 02:00pm IST. The Topcoder Collegiate Champion and the Runner Up in the Onsite Finals will get awarded with 25000 INR and 10000 INR in cash respectively along with TCC Trophies.

Apart from the obvious SRM competition, there will be talks by Topcoder members and a lot of fun games and puzzles around the arena. For all the algorithm enthusiasts, there will also be a talk on algorithms from a very well known Topcoder member who has represented India at ACM ICPC World Finals!

Looking forward to meet some of you in person at the onsite round! Good luck and have fun.

 
 
 
 
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3 years ago, # |
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"Note that you must be a student studying in an Indian university to be eligible for the next round."

Does that mean anybody can play in this round, even if they are not eligible to go to the next round?

I notice on the Topcoder calendar this is posted as "TCC India + Fun SRM (Open Invite)" which sounds like anyone can play.

Thanks for any clarification.

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    3 years ago, # ^ |
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    The "Fun SRM" is a parallel round with the TCC India Qualification SRM. Anyone can take part in it.

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      3 years ago, # ^ |
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      Is it going to be rated?

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        3 years ago, # ^ |
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        TCC Qualification and Onsite Rounds are rated. The "Fun SRM" is unrated.

        Edit : The fun round is also rated now! Sorry for any inconvenience.

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3 years ago, # |
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** deleted **

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3 years ago, # |
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How to solve medium problem?

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    3 years ago, # ^ |
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    dp(node , is the topmost node free or not) , can be calculated in O(N) but then O(N^4) passes anyways so you can do that as well.

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    3 years ago, # ^ |
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    Split the tree into minimum number of disjoint paths (using DP). Ans = no of paths-1

    DP(v) returns a pair (A, B), where A = no of disjoint paths in subtree of v with v as endpoint of some path (so that path can be extended), and B = min no of disjoint paths in subtree

    Final ans = DP(root).B-1