I think it starts at 2:00 UTC, since in the official schedule there is an entry like "Contestants enter contest area", from 1:20 UTC to 1:40 UTC

Good luck Hiasat :)

This is the first time i am upvoting this question :))

Respect to the originals.

http://codeforces.me/blog/entry/44943#comment-294851

Twitch is not really blocked. It's just slow and maybe sometimes unstable, but that's true for almost all foreign sites. Also I think non-English streamers generally don't like to use Twitch.

Все ведь помнят, что "континент" — это не "материк", и их можно делить по-разному =)

On http://online.acmicpc.org you can try this year's problems on a copy of the our submission system. It will be available 15 to 60 minutes after the start of the contest.

?detaR tI sI

I was just trying to rickroll everyone :(

Codeforces: Y U NO LET ME RICKROLL?!

A relatively simple voronoi diagram algorithm is just n^2 log n using a stack for everyone point and sorting the bisecting lines around it. I just used my Delaunnay triangulation code to build the voronoi diagrams in n log n then clip the polygon with it.

No more Moscow Secondplace University:)

Вот она! И в пост добавлю, спасибо! https://www.youtube.com/watch?v=Q2tKpmsmhgo

Nice catch

What kind of issues happened?

Congrats UCF!!!

The ones nobody solved. :(

Teams are asked to be happy / celebrate something for a few seconds. It's supposed to be shown in the broadcast when a team solves a problem.

Porto, Portugal

There are no ups without falls..

I assume Per and Onufry will post their usual solution sketches soon, and hopefully ICPCNews will post videos. I put my solutions from test-solving up here, but note that they're not always optimal (or even good) algorithms. :)

The solution sketches and analyst solution videos have been posted.

Александр, я с удовольствием готов выслушать Ваши претензии, но агрументированные, чтобы я мог что-то подправить. Лучше сделать это в личку.

.

Я просто хочу отметить, что трансляции, в особенности неанглоязычные, предназначены не для участников, а для тех, кто хочет информированно поболеть. Если убрать из трансляции все эмоции, то она станет бесконечно скучна

Лично мне в трансляции не понравилось скорее то, что ведущий активно поддерживает команды из региона Northern Eurasia, а успехи других команд, наоборот, комментирует так, как будто это что-то нежелательное. Мне кажется, что ведущий трансляции должен быть более нейтральным.

Ну и называть в официальной трансляции мирового чемпионата Виталика Виталиком как-то несерьёзно.

I would suggest looking at this.

There are two ways to construct the tree that I know of.

The first way: take a node with degree one and connect it to the node with the highest degree at the moment. Decrease the degrees accordingly and repeat n - 1 times. There are two observations that make this possible. The first is that if we are given that a tree can be built, then there is a node with a degree of one. That's because if there were no nodes with degree one then the sum of degrees would be at least 2n while any tree has degree sum of 2n - 2. The second is that after connecting that edge, we have a new degree sum of 2n - 4 = 2(n - 1) - 2 meaning that we reduced the problem to building a tree on n - 1 nodes which we assume can be done (the inductive argument should be made the other way round but I will omit it since it is of little relevance).

The second way: sort the nodes in decreasing order of degree. Make the first one the root. Maintain a list of "available" nodes that can still have neighbors. On each step take the next node and connect it to a node from the list. Reduce the degrees appropriately, remove a node from the list, and add a new node to the list if necessary. It's not hard to see that after each step the graph on the processed nodes is indeed a tree. We are left to show that on each step the list of "available" is non-empty. Assume it is empty, that means that on the previous step we added a node with a degree of one since, otherwise, that last added node would have been added to the list of "available" nodes after reducing its degree by one. Since we sorted the degrees, that means the rest of the nodes that have not been processed have a degree of one as well. Let's say that we processed k nodes so far. Since those k nodes form a tree and there are no "available" nodes among them, the original sum of their degrees is 2k - 2. The sum of the rest n - k nodes is n - k because each of them has a degree of one. So the total degree sum is n + k - 2. Observe that n + k - 2 < 2n - 2 whenever k is not n. That means that the only time that the list of "available" nodes becomes empty is when k = n meaning that we have processed the whole tree.

The first solution, in my opinion, is easier to understand. However, the second one is easier to code.

My solution: Solve the case N ≤ 2 manually. Otherwise there is at least one non-leaf vertex and at least two leaf vertices. Take all non-leaf vertices and form a chain. Then, to each vertex in a chain, connect enough leaves to it to increase its degree.