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Zlobober's blog

Opencup: Grand Prix of Baltics
By Zlobober, history, 9 years ago, translation, In English

Opencup: GP of Baltics has just finished. We are curious, what are the normal solutions for A, B and F?

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9 years ago, # |
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F: Dynamic programming d[mask] = (minimal number of subsets one can split mask into, minimal sum in the last subset).
A: Define convolution conv(d)[mask] = sum {d[x] | x is submask of mask}. One can calculate conv and conv^-1 in O(n2^n). Let clique[mask] = 1 if mask is clique, 0 otherwise. One need to check conv^-1(conv(clique)^a * conv(antilclique)^b)[full mask] != 0 -- then one can split graph in no more than a cliques and b anticliques. This can be done in O(n2^n) overall.

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    9 years ago, # ^ |
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    Can you explain in detail, why "conv^-1(conv(clique)^a * conv(antilclique)^b)[full mask] != 0"?

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

      "conv^-1(conv(clique)^a * conv(antilclique)^b)[full mask]" is a number of ways to choose (not necessarily distinct, possible intersecting) a cliques and b anticliques covering entire graph (i.e. all vertices).

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    9 years ago, # ^ |
      Vote: I like it +18 Vote: I do not like it

    For the fixed a and b, I can calculate conv^-1(conv(clique)^a * conv(antilclique)^b) in O(n2^n) easily. But how to make it O(n2^n) overall?

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      9 years ago, # ^ |
        Vote: I like it +13 Vote: I do not like it

      You can use two properties of the problem to speed it up: 1. in a row (or a column) ones will form an interval 2. you only need last element of conv^-1(...), not the entire result

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      9 years ago, # ^ |
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      Minor addition to ikatanic answer: last element (in fact, any single element) of conv and conv^-1 can be calculated in O(2^n).

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    9 years ago, # ^ |
      Vote: I like it +13 Vote: I do not like it

    Could you please give some more details or your code for F.

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9 years ago, # |
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Can some1 describe a solution for C in details more or less?

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9 years ago, # |
← Rev. 2   Vote: I like it +8 Vote: I do not like it

Is it possible to see solutions or editorial of this contest ?

If yes, please share link.

I want to see solution of 'F'. problem link is here. Thanks in advance

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