We will hold UNIQUE VISION Programming Contest 2023 New Year (AtCoder Beginner Contest 287).

- Contest URL: https://atcoder.jp/contests/abc287
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20230128T2100&p1=248
- Duration: 100 minutes
- Number of Tasks: 8
- Writer: Kodaman, m_99, PCTprobability, kyopro_friends, nok0
- Tester: yuto1115, physics0523
- Rated range: ~ 1999

The point values will be 100-200-300-400-500-500-600-600. We are looking forward to your participation!

I can't seem to figure out why my F gets TLE. I used the same trick as the editorial. Link.

Is it some cache-friendly stuff that I am missing? Or is there some other stupid thing that I am doing in my code.

It is due to modulo operator most probably, try using Mint.

That doesn't seem likely to me, because modint is just a class-level abstraction for modular operations that makes it less likely to make overflow errors. I don't think that would actually make it less costly (in fact, because of the wrapper, I would guess it makes the operations slightly more costly).

I did try it though: it still TLEs on most of the tests: Link

Having lots of small-sized vectors is bad. Simply replacing all your 2-sized vectors with

`array<int, 2>`

makes it run in 639ms. Submission.Thanks for your insight. I saw tourist casually use multiple 3D vectors in his streams, and so thought that it was perfectly fine (although later he did change it to arrays after getting a TLE xD)

Could have solved G for the first time in contest today. But I am so bad at implementation :((

BTW today's contest seemed a lot less

mathy. Good job!I have an $$$O(N \log N + \sum |S_i|)$$$ solution for E. Observe that the string having maximum LCP with $$$S_i$$$ would be, the one before it and the one after it, when we sort the array $$$S$$$. So, we manage two multisets $$$L$$$ and $$$G$$$, each sorted non-decreasing and non-increasing. Now the rest of the solution is basically calculating the LCP value. Submission here.

We do not need two multisets. We can just put the strings in an array, sort the array and for every string: check the LCP of the two neighboring strings. Submission here.

Yes, I knew, I was too lazy to do some work with indices/iterators.

After I continually get WA in C, I got impatient and finally got -58 in this round.

After seeing the editoral, I thought my mistake is quite stupid.

:(

Some people falsely got AC for C.

Problem E can be easily solved using Trie with a complexity of $$$O(\sum {|S_i|})$$$. Here is my solution

True. I was also going the same direction but I realized the max LCP should be largest for the closest string lexicographically greater or smaller than the given string.

This was faster to code but definitely slower than the trie approach.

Decided to use Trie so that I could practice implementing it.

Did it for the first time in several months, lol.

The set-based implementation would've been a lot faster to code.

I passed E with Time Complexity $$$O(n\log^{2} \sum S_i)$$$. Maybe the TL limit should be a little tighter.

I use double hash to solve E. My idea is converting all possible prefixes using double hash into a map

For each string I do binary search the answer tp check whether a certain prefix exist in the map

Thanks for the problemset! I love E and G (can't solve G on contest but the double fenwick tree + compression is cool!)

Thanks for the round. Really one of my favorite rounds which includes diverse problems in different topics.

btw, it's my first rated round.

@chokudai I think Task C has weak test cases. I have dmed you the details.

Solved till E for the first time. Although first two were way easier than div2 A.

Can anyone explain F?

G is solvable using treap. Submission

For problem F, during contest I have almost figured out the solution but failed the complexity analysis part. I could not convince myself that it is O(N^2) rather than O(N^3), until reading the editorial and realize that maybe it is similar to this problem https://codeforces.com/contest/581/problem/F.

I have seen this problem/idea during virtual participation recently but missed such observation during today's contest. Although it is a pity, still thanks to writers for providing such invaluable problems.

Explanation for $$$F$$$ (Same Editorial's idea but trying to make it more clear):

Let $$$dp[i][is\_chosen][cnt]$$$ be the number of subsets with $$$cnt$$$ connected components over the subtree rooted at node $$$i$$$. $$$is\_chosen$$$ is $$$0$$$ or $$$1$$$ and represents whether the subsets include $$$i$$$ or not. After processing the $$$k^{th}$$$ child of $$$i$$$, $$$dp[i]$$$ should have the number of subsets over $$$i$$$ and the subtrees rooted at its first $$$k$$$ children.

When processing the $$$(k+1)^{th}$$$ child (let it be $$$j$$$), all we have to do is to merge $$$dp[j]$$$ to $$$dp[i]$$$, i.e., iterate on every $$$cnt_i$$$-$$$cnt_j$$$ pair and add their contribution $$$dp[i][is\_chosen_i][cnt_i]\cdot dp[j][is\_chosen_j][cnt_j]$$$ to $$$merged\_dp[is\_chosen_i][cnt_i+cnt_j-(is\_chosen_i\ \&\&\ is\_chosen_j)]$$$.

After processing the $$$(k+1)^{th}$$$ child, just assign $$$merged\_dp$$$ to $$$dp[i]$$$. Note that we subtract $$$(is\_chosen_i\ \&\&\ is\_chosen_j)$$$ because if both $$$i$$$ and $$$j$$$ are chosen, they are merged into $$$1$$$ component. We need to take care here because the expression $$$cnt_i+cnt_j-(is\_chosen_i\ \&\&\ is\_chosen_j)$$$ can be $$$-1$$$.

Proof that complexity is $$$O(N^2)$$$:

Observe that when we iterate on every $$$cnt_i-cnt_j$$$ pair, we can imagine this (in terms of iterations count) as iterating on every pair of nodes where the first node is from the currently merged subtrees and the second node is from the new subtree to be merged. Hence, every pair of nodes will be considered only once, i.e., at their lowest common ancestor. Since we have in total $$$\dfrac{N\cdot (N-1)}{2}$$$ pairs, the total number of iterations is bound by that count.

Submission.

Does anyone know how the test cases for E is generated?

I hashed every prefix using a mod of 10^9+7 with a random base (tracking the hashes of different lengths separately). I thought because of the low number of test cases on atcoder, that would be enough. But surprisingly I still failed 3/33 test cases. It passed when I switched to a different (and larger) mod.

give me test cases where

this will give the wrong answer.

for problem C

The second diagram in the editorial

why this one

try this testcase:

answer:No

You are right; now I'm thinking it was too stupid to ask. Thanks anyway.