Блог пользователя arsijo

Автор arsijo, 5 лет назад, По-английски

Hi everyone,

I am happy to announce that the 2019 Southeastern Europe Regional Contest will take place on October 19 in Bucharest and Vinnytsia at 10 am UTC+3. The link to the live results will be published here after the start of the competition.

Moreover, in the following day, there will be the Grand Prix of Southeastern Europe based on these problems at 8 am UTC. If you do not participate in it, you will be able to take part in online mirror, which will be available at the same time.

Because of the Grand Prix, we are asking the official contestants not to discuss the problems in public.

After it, we will upload the contest to the gym and publish the editorial.

Good luck!

UPD. SEERC standings.

UPD2. Congratulations to the winners!

Place Team Name Contestant 1 Contestant 2 Contestant 3 Problems Penalty
1 [UNIBUC] ScrambledEggs bicsi theodor.moroianu livlivi 8 1001
2 KhNURE_NRG BigBag Mustang98 shanin 8 1077
3 RAF Penguins Pajaraja milisav allllekssssa 8 1401
4 KhNU_DOROGA* Stroustrup kilt_01 13022001 7 976
5 KNU_MaxFun Adalbert Barichek Markellonchik 7 1027
6 UNAGI osmanorhan ikbal ozcelik 7 1498
7 LNU Jackals Danylo99 mshcherba BohdanPastuschak 6 508
8 UBB_HakunaMatata Bodo sebinechita Cozma_Tudor 6 719
9 KNU_Kektus* danya.smelskiy iSlava PashaChemerys 6 777
10 [UAIC] VYE Juve45 cristian1997 denis2111 6 824

UPD3. Online mirror will be available here. PS. You should join the group first. The mirror will start after the Open Cup.

UPD4. Editorial is here.

UPD5. The contest is available in the gym.

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5 лет назад, # |
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And this GP clashes with cf div 1, right?

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5 лет назад, # |
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In VYE team, the third correct person is cristian1997

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5 лет назад, # |
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I am positively surprised by the quality of problems. I associated SEERCs with very bad problems, but they were actually quite good (except tedious K), I would say CFH were AtCoder quality problems

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5 лет назад, # |
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How many teams qualified to WF from SEERC ?

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5 лет назад, # |
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There is another way to come to solution for problem F.

Let's count the number of times first player can win in subtree, assuming black vertices remain black -- $$$f(v)$$$. Each player can refuse to move and lose.

If all sons are losing than first player chooses the root and second player then refuses to move. So in such case $$$f(v)=1$$$.

If there are at least 1 winning son, first player wins in that son forcing second player to move the root and then first player moves again to the same or another winning son. All other times second player can't move to the root, so first player just wins all sons, so $$$f(v)=\sum{f(to)} - 1$$$