dalex's blog

Gym — XI Samara Regional Intercollegiate Programming Contest
By dalex, 7 years ago, translation, In English

Hello everyone,

At March 11 there was an annual programming contest in Samara University, and yet again we copy it into Codeforces Gyms. The gym contest takes place on Sunday, March 18, at 11:00 MSK.

Third time in a row it is a personal contest. So we ask everyone to participate solo. The contest is the most interesting for blue and violet coders.

And as usual,

The list of our previous contests
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7 years ago, # |
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Can I register with my team? Or only individual contestants?

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    7 years ago, # ^ |
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    You can, but people on the onsite contest played individually.

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7 years ago, # |
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Will there be an editorial after the contest. It becomes really hard to get solutions for all the problems from the comments for beginners like me.

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7 years ago, # |
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I enjoyed the contest, the problems were really interesting. Can I see the case I'm failing at on problem M?

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    7 years ago, # ^ |
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    This test has n = 3.

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      7 years ago, # ^ |
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      I'm like brute forcing, so maybe it's some logical bug.

      Can you share the full test? :)

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7 years ago, # |
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Can someone share his code for C.

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    7 years ago, # ^ |
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    Code

    I'm not sure how clear it is but I can explain anything if it's unclear.

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      7 years ago, # ^ |
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      Could you please explain how your code works. This is my Code. The Idea is same as yours I guess but its giving wrong answer on Test 11.

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        7 years ago, # ^ |
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        I start by sorting the ranges by their end points. Then as I'm processing the end points, ill greedily build my set.

        Basically, if by the time I get to the end of some range and there's no number in my set (that's the ceiling() call) that belongs in it's range, then I'll add that range's endpoint to the set.

        This is always optimal because I'm going in increasing order of endpoints so every time I add a number to my set, it's as large as possible so it's most likely to fit in the other ranges (who all have end points greater than the current one).

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          7 years ago, # ^ |
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          Why I sort the ranges by their start points then I got wrong answer ? But sort the ranges by their end points will get AC ?

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7 years ago, # |
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Any hint for problem H or L???

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    7 years ago, # ^ |
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    Problem H:

    BFS off of every monster tile and only go a maximum of d distance away. Any tile that was touched by a monster's BFS is now no longer usable. Now BFS from your starting point and do regular shortest path in a grid but making sure to ignore cells touched by the monsters.

    Problem L:

    For every position i in the string, find the next position of every letter in the alphabet. Now keep some kind of queue and find the next position of whatever letter is queried (or pop the back of your queue). If the next position of the queried letter doesn't exist, the answer is no otherwise it's yes.

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      7 years ago, # ^ |
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      So simple BFS's were enough in H? I thought that with n*m up to 4*10^10 that would take too long, even just to calculate 'bad' cells

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      7 years ago, # ^ |
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      For H，first i DFS for each 'M' finding which grid can't reach, then BFS for shortest path, but wrong answer on test 20，i really wanna know what test 20 data is,thx~

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        7 years ago, # ^ |
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        Go here

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        6 years ago, # ^ |
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        Yeah I too got a WA on test 20 , but I don't know why this approach gives WA with dfs

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          6 years ago, # ^ |
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          Test 22 is here, most probably you can't pass it too.
          Test 20 is the same but with other parameters:

          Spoiler
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            6 years ago, # ^ |
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            But can you help me out why, am I getting a wrong answer. Is doing dfs to find cells at distance d wrong?

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              6 years ago, # ^ |
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              Yep, it's definitely wrong. You may visit some cell with distance X <= M, and then, when you will try to visit it again with distance Y < X <= M, you will ignore that edge because cell already visited. So the fix is change visited array type from boolean to int (shortest distance from M to that cell), but in that case I think you may get stuck with TLE.

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                6 years ago, # ^ |
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                Yeah I encountered a TLE, with bfs

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                  6 years ago, # ^ |
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                  1. Don't run bfs from each M separately (run normal bfs, just put all 'M' together to the queue with H=0)
                  2. Try to replace HashMap by array
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                  6 years ago, # ^ |
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                  can someone explain why this code gives MLE (the queueis full ) i used bfs from all M then one bFS from s for path length https://pastebin.com/ip8Q1hH8 solved : I have to mark the node visited than added to the queue not the inverse

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7 years ago, # |
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What's output of problem E for this test case: abaca aabac

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7 years ago, # |
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What is the test 48 on F?

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    7 years ago, # ^ |
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    During the contest this test helps me:

    4
3 2 3 4
1 4
1 4
0

    Answer is NO

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7 years ago, # |
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Any hints on how to approach I? My approach can't seem to get past test 12.

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    7 years ago, # ^ |
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    I used this approach:

    Define a procedure findRoot(vertices, leafs, depth) for search a tree root among vertices, inside it we need to find two leafs from a different subtrees (relative to root). It can be managed in O(size(vertices)), then you can easily find a root (it is a vertex, that dist(leaf1, v) = dist(leaf2, v) = depth). Then you need to run findRoot recursively from left/right subtree.

    So, each findRoot procedure takes 2 * size(vertices) queries (2 * nlogn in total) and additional queries at start to find leafs.

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      7 years ago, # ^ |
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      That makes a lot of sense. Thank you!

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      7 years ago, # ^ |
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      Hi, I can't figure it out how do you manage to classify vertex according to left and right subtree for the recursive call. Can you explain that part? Because you pass vertices and leafs parameters that, I guess, represent vertices and leafs of the current subtree.

      Thanks

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        7 years ago, # ^ |
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        You have two leafs from different subtrees, so lets call them left & right. So to classify arbitrary vertex we need to compare dist(left, vertext) and dist(right, vertex) — if dist from vertex to left vertex smaller then vertex in a left subtree and vice versa.

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      7 years ago, # ^ |
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      Could you show me the code? I try many many ways but i still can't figure it out.

      Thanks.

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        7 years ago, # ^ |
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        It's very easy to include interactor in the solution and test it locally. All tests for this problem are just trees with random-shuffled vertex indices (there are no any special cases). So it doesn't seem you "tried many many ways".

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          7 years ago, # ^ |
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          That's because I make too many queries then I get a wrong answer,but i have no idea how to reduce the query though I know my ways can restore the tree.

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            7 years ago, # ^ |
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            Spoiler
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7 years ago, # |
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What's test 22 on H ?

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    7 years ago, # ^ |
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    Test 22 generator
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7 years ago, # |
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The game .C.O.N.T.E.S.T: Unexpected Behaviour from Problem K "Video Reviews" does not exist. But recently our gamedev team Veslo Games released the game .T.E.S.T: Expected Behaviour (from creator of Codeforces Simulator!) to the Steam Early Access. If you like puzzles, consider checking this out!

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7 years ago, # |
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is there a algorithm about problem M？or it's just a simulation？

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    7 years ago, # ^ |
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    Spoiler, M
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7 years ago, # |
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could anybody help me with the problom I, I can't pass the test 5

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    7 years ago, # ^ |
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    This test has n = 7 and random-shuffled vertices.

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      7 years ago, # ^ |
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      thanks a lot,but i'm stuck in test 12 now, trust me,i will figure it out :( lol

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7 years ago, # |
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I am stuck on problem H 8-th test . Can I see the case I'm failing ?

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7 years ago, # |
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Can anyone provide any hints for problem K ,video reviews? thanks.

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    7 years ago, # ^ |
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    Main hint: binary search

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      7 years ago, # ^ |
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      thanks for replying.binary search i figured out but was not able to come up with a predicate.thanks.

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        7 years ago, # ^ |
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        Suppose we have X, and we need to check if X is enough to get m reviews. You need to process the bloggers from left to right. There are two options:

        • blogger is already interested in the game — just increase reviews count
        • blogger isn't interested in the game — you need to convince him (if already convienced count < X) and increase reviews count.

        It can be easily proven that greedy algorithm works.

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7 years ago, # |
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What is the test case 50 in problem F? I got stucked there :(

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    7 years ago, # ^ |
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    F, test 48
    F, test 50
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      7 years ago, # ^ |
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      Thanks, I don't know why my aproach was wrong. At the end I changed my strategy and get AC.

      Do you have any hint for problem D? I belive that a maxflow algorithm could said if it is possible or not, but I can't construct the solution

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        7 years ago, # ^ |
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        Certificate in D
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7 years ago, # |
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What is the test case 52 in problem F? I got stucked there.

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    7 years ago, # ^ |
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    The same as test 48 (see my comment above), but with different random seed

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6 years ago, # |
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Can someone that give hints about questions D, G and I ? I'm stuck on those.

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6 years ago, # |
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May I get some help about 112 No. test case of problem No. F ???

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    6 years ago, # ^ |
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    The same as test 48 and 52 (see my comment above), but with different random seed

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6 years ago, # |
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Problem H. Safe Path

WA on Test Case 124 :(

I am stuck.

Here's My Code:

https://codeforces.me/gym/101755/submission/48191879

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6 years ago, # |
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Beautiful problems, solved all, recommend to everybody

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5 years ago, # |
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Any hints of for problem D?

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