dario2994's blog

SWERC 2021-2022
By dario2994, 3 years ago, In English

The Southwestern Europe Regional Contest will take place on the 23rd of April. It is the ICPC regional contest (i.e., the winning teams will advance to the ICPC World Finals) for teams from France, Israel, Italy, Portugal, Spain, and Switzerland.

The mirror contest SWERC 2021-2022 - Online Mirror (Unrated, ICPC Rules, Teams Preferred) will be held on Codeforces at Apr/24/2022 14:05 (Moscow time) and will last 5 hours.

The mirror contest will contain all the problems from the official competition plus some additional problems.

I am the chief judge for the competition and I want to thank:

I invite you to participate in the contest and I hope that you will like the problems.

On the difficulty
The contest features problems with difficulties from div2A to div1F; so anyone can find something at their level.

Many teams without much experience participate in SWERC, so the problem set should be enjoyable also for div2 contestants. On the other hand, solving all the problems should be challenging even for the strongest teams in the world: the MIT team did not AK in 5 hours.

On the beauty

Rules

  1. The contest is unrated, so your codeforces rating will not be affected.
  2. The scoring is ICPC-style: teams are first sorted by number of problems solved, then the time-penalty is used as a tie-break. An incorrect submission gives a 20 minutes penalty.
  3. We encourage participation as a team.
  4. If you are participating in a team, we encourage you to use only one computer for coding the solutions (as in an ICPC contest). Regarding using templates and copy-pasting code: feel free to do it.
Rationale of rule 4.

UPDATE: We hope you liked the problems, here is the editorial for all the problems in the mirror: https://mirror.codeforces.com/blog/entry/102042 .

UPDATE: Congratulatins to the all the participants of the onsite contest and in particular to the two gold medal winning teams, both solving 10 problems (and with a very similar penalty time):

  1. Raw Pots -- Harbour.Space University
  2. TAU++ -- Tel Aviv University

And congratulations also to the four teams who managed to solve all the problems in the mirror:

  1. tourist, ksun48
  2. jiangly
  3. Merkurev, KAN, Um_nik
  4. djq_cpp, hehezhou, jqdai0815
  • Vote: I like it
  • +482
  • Vote: I do not like it

Comments Comments (43)
| Write comment?
»
3 years ago, # |
  Vote: I like it +11 Vote: I do not like it

orz (i wonder why no one else has commented this so far)

→ Reply
»
3 years ago, # |
← Rev. 2   Vote: I like it +23 Vote: I do not like it

Why the registration is closed at once?

Edit: Now it's open. Thanks!

→ Reply
»
3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Is the online mirror private for certain users?

→ Reply
»
3 years ago, # |
  Vote: I like it +18 Vote: I do not like it

why some participants are colored red for this contest??

→ Reply
  • »
    »
    3 years ago, # ^ |
      Vote: I like it +43 Vote: I do not like it

    They registered twice (probably because they registered on their own, then another team registered them). Just unregister one of which and it should be fine.

    It won't prevent them from joining the contest either way, but last I heard, if they submit, a random entity out of the two will get counted for submission.

    → Reply
»
3 years ago, # |
  Vote: I like it +2 Vote: I do not like it

clash with google codejam 1B

→ Reply
  • »
    »
    3 years ago, # ^ |
      Vote: I like it +9 Vote: I do not like it

    By 5 minutes??

    → Reply
  • »
    »
    3 years ago, # ^ |
      Vote: I like it +48 Vote: I do not like it

    Yes, we are aware of it. We think that the last 5 minutes of a 5 hours mirror and/or the first 5 minutes of google code jam round 1b are not super important and it's acceptable if they overlap.

    → Reply
»
3 years ago, # |
← Rev. 3   Vote: I like it -177 Vote: I do not like it

nice

→ Reply
  • »
    »
    3 years ago, # ^ |
      Vote: I like it +81 Vote: I do not like it

    You are absolutely right, this contest is completely useless for you.

    → Reply
»
3 years ago, # |
  Vote: I like it +23 Vote: I do not like it

Good chance for training on online contest because rated will not be affected

→ Reply
»
3 years ago, # |
  Vote: I like it +10 Vote: I do not like it

I am excited to participate with my friends today in this contest.

→ Reply
»
3 years ago, # |
← Rev. 3   Vote: I like it 0 Vote: I do not like it

Is there a live scoreboard of the official competition?

UPD: Found it: https://scoreboard.swerc.eu

→ Reply
  • »
    »
    3 years ago, # ^ |
      Vote: I like it +34 Vote: I do not like it

    Rule 5. If you want to participate "competitively" in the mirror, refrain from opening the scoreboard so that you don't obtain any information on the difficulty of the problems.

    → Reply
»
3 years ago, # |
← Rev. 2   Vote: I like it +54 Vote: I do not like it

Problem N be like :

Coincidentally, she notices that no two contestants have the same age, and that everyone is between 1 and $$$n^2$$$ years old.

Contestants with age 2,250,000 :

→ Reply
»
3 years ago, # |
  Vote: I like it +1 Vote: I do not like it

Hi, I'm new to this type of contests and I have trouble trying to register: https://streamable.com/jjl827 This was happening since I tried registering from before the contest started. Can I have some advice on what I can do for future contests please?

→ Reply
»
3 years ago, # |
  Vote: I like it +10 Vote: I do not like it

good round, really enjoy this round

thank you for the problem

problem D H is cool and O is also nice (i guess the traditional way works)

M is also a good (not sure because of not implement)

how solve to B C E F G J K L N?

→ Reply
  • »
    »
    3 years ago, # ^ |
      Vote: I like it +10 Vote: I do not like it

    M is very straightforward, just put maxR 'R' and maxW 'W'. how to solve D? My thought is remove all occurence of AA, BB, CC and replace AB with BA, CB with BC, then check if the two string is same. However it fails. What did I miss?

    → Reply
    • »
      »
      »
      3 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      After AB->BA replaces some new occurrences of AA or BB may appear. But if you will repeat these 2 sets of substitutions until none could be applied any more, then, I guess, this will be correct.

      → Reply
      • »
        »
        »
        »
        3 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        I did it recursively. Let me wait for editorial or failure cases.

        → Reply
    • »
      »
      »
      3 years ago, # ^ |
      ← Rev. 2   Vote: I like it 0 Vote: I do not like it

      That should work. In a simpler O(N) constructions: basically send all the B's to the start (and remove BB), and use a stack to pair up AA and CC. I don't know how to prove it's correct though.

      → Reply
  • »
    »
    3 years ago, # ^ |
    ← Rev. 2   Vote: I like it +8 Vote: I do not like it

    K:

    Note that due to continuity and such arguments, it is not hard to verify that the sum of lengths can be made the same for all three triangles when you reach the minimum $$$r$$$. Let the given points be $$$A, B, C$$$. Construct points $$$A'_i, B'_i, C'_i$$$ with $$$i \in \{1, 2\}$$$ such that $$$BCA'_i$$$ etc. are all equilateral. Let's call a triangle good if the minimum sum is achieved at a vertex, and bad otherwise. If we choose an optimal residence point $$$D$$$, then there are at least two triangles among $$$DBC, DCA, DAB$$$ which are both good or both bad, since a triangle is good iff one angle of the triangle is $$$> 120^\circ$$$. In the case they are good, say $$$DAB$$$ and $$$DCA$$$ are good, we just need to ternary search for the best point on the corresponding perpendicular bisector of $$$BC$$$ (do this for all permutations to find the answer in this case). In the case they are bad, say $$$DAB$$$ and $$$DCA$$$ are bad, note that we have $$$DB' = DC'$$$ where $$$B', C'$$$ are on the opposite sides of $$$CA, AB$$$ as $$$B$$$ and $$$C$$$. So we should ternary search for the answer on the perpendicular bisector of $$$B'C'$$$ (and do it for $$$A'B'$$$ and $$$C'A'$$$ as well).

    Note that whenever we said that we need to do a ternary search, there are some more constraints to be satisfied. We note that the problem is to find a global minimum, and since the function is convex, we should just evaluate the answer for each case naively while ternary searching (without paying attention to such constraints). For instance, rather than using $$$DA'$$$, we can use $$$\max(DA', DA' ')$$$.

    → Reply
»
3 years ago, # |
  Vote: I like it +46 Vote: I do not like it

Now that the contest is over, why are we not able to submit to the problems in practice mode?

→ Reply
»
3 years ago, # |
  Vote: I like it +1 Vote: I do not like it

Is there any editorial?

→ Reply
»
3 years ago, # |
← Rev. 2   Vote: I like it +21 Vote: I do not like it

..

→ Reply
»
3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Problem I is a much easier version of USACO21DEC Silver Closest Cow Wins. In this problem $$$p_i$$$ is fixed and $$$N=1$$$.

→ Reply
»
3 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Why still can't view the submissions?????

→ Reply
»
3 years ago, # |
  Vote: I like it +1 Vote: I do not like it

will teams' pics be added when clicking on them on the scoreboard?

→ Reply
  • »
    »
    3 years ago, # ^ |
      Vote: I like it +11 Vote: I do not like it

    Yes, in the next few days the swerc.eu website will be updated with a copy of the scoreboard that includes team's photos.

    → Reply
»
3 years ago, # |
  Vote: I like it +14 Vote: I do not like it

When will we know who's invited to WF (now that all european regionals finished)?

→ Reply
»
3 years ago, # |
  Vote: I like it -8 Vote: I do not like it

MikeMirzayanov, it has been more than 2 weeks since this contest, please update problem ratings. Thanks.

→ Reply
»
3 years ago, # |
  Vote: I like it +19 Vote: I do not like it

I just came here to say how amazingly clever problem N is

→ Reply