Тренировка — XI Самарская областная межвузовская олимпиада по программированию

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4	Radewoosh	3633
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7	ecnerwala	3446
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3	maomao90	162
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7	awoo	157
8	TheScrasse	154
9	Dominater069	153
9	nor	153

Привет.

11 марта в Самарском университете состоялась ежегодная олимпиада по программированию, и мы снова выкладываем ее в тренировки Codeforces. Тренировка пройдет в воскресенье, 18 марта, в 11:00 MSK.

Этот контест уже третий год проводится личным. Поэтому просим всех тоже участвовать лично. Наиболее интересно должно быть фиолетовым и синим участникам.

Ну и как обычно,

Список предыдущих наших контестов

Комментарии (69)

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karamkontar

7 лет назад, # |

+13

Can I register with my team? Or only individual contestants?

→ Ответить

dalex

7 лет назад, # ^ |

You can, but people on the onsite contest played individually.

CaptainWayne

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.

Osama_Alkhodairy

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

This test has n = 3.

I'm like brute forcing, so maybe it's some logical bug.

Can you share the full test? :)

Can someone share his code for C.

Ahmad

Code

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

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.

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).

linh334qthd

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 ?

← Rev. 2 →

Any hint for problem H or L???

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.

didenko.ilia

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

n * m ≤ 2 * 10⁵

Oh.. I guess I can't read :(
Thanks, now it's a lot more clear

cucrui

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~

Go here

its_ulure

6 лет назад, # ^ |

-8

Yeah I too got a WA on test 20 , but I don't know why this approach gives WA with dfs

Test 22 is here, most probably you can't pass it too.
Test 20 is the same but with other parameters:

Spoiler

But can you help me out why, am I getting a wrong answer. Is doing dfs to find cells at distance d wrong?

pitfall

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.

← Rev. 3 →

Yeah I encountered a TLE, with bfs

Don't run bfs from each M separately (run normal bfs, just put all 'M' together to the queue with H=0)
Try to replace HashMap by array

J4TR

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

Timurovich

Задачи в ваших контестах очень интересные и разнообразные! Продолжайте в том же духе!

What's output of problem E for this test case: abaca aabac

The output is "NO"

Kerim.K

What is the test 48 on F?

During the contest this test helps me:

Answer is NO

Any hints on how to approach I? My approach can't seem to get past test 12.

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.

That makes a lot of sense. Thank you!

Silence_for_Melody

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

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.

imoverflow

Could you show me the code? I try many many ways but i still can't figure it out.

Thanks.

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".

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.

I got it. Thanks a lot.

What's test 22 on H ?

Test 22 generator

n = 5000;
m = 40;
d = 5;
a.assign(n, string(m, '.'));
a[0][m/4] = 'S';
a[0][m/2] = 'M';
a[0][3*m/4] = 'F';
for (int i = d + 1; i < n - d - 1; i++) {
    for (int j = d + 1; j < m - d - 1; j++) {
        a[i][j] = 'M';
    }
}

craus

+29

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!

AnchorCat

is there a algorithm about problem M？or it's just a simulation？

Spoiler, M

could anybody help me with the problom I, I can't pass the test 5

This test has n = 7 and random-shuffled vertices.

thanks a lot,but i'm stuck in test 12 now, trust me,i will figure it out :( lol

xskhirtladze

I am stuck on problem H 8-th test . Can I see the case I'm failing ?

Test Case 8

ucrop

Существует ли разбор контеста?

abaddon666

Can anyone provide any hints for problem K ,video reviews? thanks.

Main hint: binary search

thanks for replying.binary search i figured out but was not able to come up with a predicate.thanks.

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.