megalodon's blog

TopCoder Open 2013
By megalodon, 11 years ago, In English

The TopCoder Open 2013 starts this saturday february 23, sponsored by Google.

For more info visit http://community.topcoder.com/tco13/algorithm/

To register visit http://community.topcoder.com/tco13/

See you on the arena. Good luck!

EDIT: The registration is now closed, as it will take only the first 2000 contestants registered for each qualification round.

EDIT: The times are now available to everyone in each timezone.

Round 1A: http://www.timeanddate.com/worldclock/fixedtime.html?msg=TCO13+Algorithm+Round+1A&iso=20130223T12&p1=98

Round 1B: http://www.timeanddate.com/worldclock/fixedtime.html?msg=TCO13+Algorithm+Round+1B&iso=20130302T12&p1=98

Round 1C: http://www.timeanddate.com/worldclock/fixedtime.html?msg=TCO13+Algorithm+Round+1C&iso=20130309T12&p1=98

For more info about the dates and times of the event, visit http://community.topcoder.com/tco13/algorithm/algorithm-schedule/

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Comments Comments (11)
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11 years ago, # |
← Rev. 8   Vote: I like it +6 Vote: I do not like it

Time at your timezone is not given anywhere. You can get it here

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    11 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    Oh yeah, thank you sir! All the times in the contest website are in UTC-5. Your link is very helpful, maybe you cand add some more for the other rounds.

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    11 years ago, # ^ |
    ← Rev. 2   Vote: I like it +2 Vote: I do not like it

    Link (not actual after edit)

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11 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

I tried to register, and it says that I already have, but I don't remember having done the registration. Is there anyway to check the list of registered people?

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11 years ago, # |
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When Can I register in the arena for round 1A?

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    11 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Registeration in the arena starts 3 hours before the contest

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11 years ago, # |
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Should we have used min-cost-max-flow in 1000?

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    11 years ago, # ^ |
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    same here: does it work? We need exactly one outgoing and one incoming edge for every vertix.

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      11 years ago, # ^ |
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      Well, really we have an assignment problem: for each cell we put a vertex to the first part and to the second part, and our edges will be the adjacency relation between neighbours (we put exactly 4 edges from each vertex of the first part). Their weight will be obviously 0 for the edge, which follows the direction of the arrow, and 1 (for all other edges).

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    11 years ago, # ^ |
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    Yeap=)

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