Arpa's blog

By Arpa, history, 3 months ago, In English,

Hi!

Thanks to all of you participates, who made this contest possible. And thanks to Lewin and Arterm, also to the great coordinator, Nikolay KAN Kalinin, zemen, white2302, and for sure MikeMirzayanov.

Test data and code solutions. It's the original packages from polygon, you can find pretests, tests, generators, validators, etc in it.

Hints

Div.2 A: Take a look at notes section.

Div.2 B: Create a circle with these points.

Div.1 B: Fix the gcd.

Div.1 C: Tag: Grundy number.

Details

Div.1 C

I want to thank my grand teacher Mojtaba moji FayazBakhsh here. Who was my teacher not only for coding but also a teacher for my life. Thanks Mojtaba! Thanks to you and all of other good teachers in the world.

Solutions

Tutorial is loading...

Author: Arpa

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Author: Arpa

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Author: Lewin

Thanks to Lewin, the writer of this tutorial.

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Author: Arpa

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Author: Arpa

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Author: Arterm

Thanks to white2302, the writer of this tutorial. I translated this tutorial to English.

Tutorial is loading...

Author: Arterm

Thanks to Arterm, the writer of this tutorial.

Tutorial is loading...

Author: Lewin

Thanks to Lewin, the writer of this tutorial.

I’d like to finish the editorial with the below poem by Hatef Esfahani:

چه کند کوه کن دلشده با غیرت عشق گر نه بر فرق زند تیشه ز رشک خسرو

Translation: What can lover (Farhad) do with the power of love? He has no choice but to hurt himself by ax because he feels jealousy to Khosrow. More information about Khosrow and Shirin story.

Good luck and see you soon ;)

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By Arpa, history, 3 months ago, In English,

Hi!

I'm honored to invite you to Codeforces Round #432, it will be held on 4th September 14:35 UTC. There will be 5 problems for each division, as usual, you have 2:30 to solve the problems. The contest was prepared by Lewin Lewin Gan, Artsem Arterm Zhuk and me.

The IndiaHacks Final Round will be held on 3rd September 12:30. Finalist must not discuss the problems after their contest.

The stories of my problems will be about Arpa, although in one problem you'll see Mojtaba Moji FayazBakhsh, my great teacher.

I'd like to thank Lewin, Artsem and myself (:P) at first, then Konstantin zemen Semenov and white2302 for testing the problems, Nikolay KAN Kalinin for helping us in moving the contest to codeforces and Mike MikeMirzayanov Mirzayanov for the great Codeforces and Polygon platforms.

The scoring distribution will be announced later.

Obviously, if you are interested in if the round is rated or not, ask in comments and get a lot of down votes.

UPD. There will be 5 problems for div.2 and 6 problems for div.1.

UPD2. Scoring Distribution: div.1: 500-1000-1250-1750-2000-2500, div.2: 500-1000-1500-2000-2500.

UPD3. Editorial is partially ready. I'll complete it soon.

Congratulations to winners:

Div.1:

1 . AngryBacon

2 . dreamoon

3 . sd0061

4 . W4yneb0t

5 . Um_nik

Div.2:

1 . miaom

2 . fzzzq2002

3 . igoodvegetableb

4 . _Lucas97 and Szymanski_w (WoW !!)

Official results

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By Arpa, history, 4 months ago, In English,

Hi!

Hints

A: Tag: Greedy!

B: Tag: Greedy!

C: For each vertex like v find exv, the expected length of their journey if they start from v.

D: Tag: Inclusion exclusion.

E: Find the maximum clique.

Solutions

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Arpa's solution: 29412123.

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Arpa's solution: 29412154.

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Arpa's solution: 29412220.

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MaGaroo's solution: 29458745.

Tutorial is loading...

Arpa's solution: 29412249.

P.S. Please notify me if there is any problem.

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By Arpa, history, 4 months ago, In English,

Hi!

Hints

A: Let Cost(k) the answer if we compress the image with k, find min Cost(k).

B: Solve the problem without first query.

C: Let the frequencies of the characters be a1, a2, ..., ak. Alice loses if and only if is odd and n is even.

D: Consider adding an extra seat, and the plane is now circular.

E: Let dpi, j be the longest path given that we've included segment i, j in our solution, and we are only considering points to the right of ray i, j (so dpi, j and dpj, i may be different).

F: We want to compute dpi, j: expected value given we have seen i red balls and j black balls.

Solutions

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Codes here.

P.S. Please notify me if there are any problems.

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By Arpa, history, 7 months ago, In English,

Hi !

About one month ago I asked Xu Han (admin of the Virtual judge) if it's possible to add my own problems (that aren't present in any online judge) to some contest in Virtual judge, and we worked together to find a way to do this, and here is it, you can now add your own problems to Virtual judge. Let's see how to do.

  1. Register on Codeforces polygon, create a problem and add statements, tests, checkers, etc.
  2. Create a mashup here.
  3. Add the problem to your mashup. Follow the instruction in this comment.
  4. From contest panel, go to "manage inventions" → "invite users" → add vjudge1, vjudge2, vjudge3, vjudge4, vjudge5.
  5. The problem is added to Virtual judge now!
  6. To add your problem to your contest, choose "gym" from judge list and add your problem using its code (contest code + problem code, i.g. 207753A).

To see an example, here is my added problem to Virtual judge: Arpa as an ARPA :P (Hard, Fixed).

P.S. Deleting problem is not available now, but Xu Han said that he will make it available soon.

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By Arpa, history, 9 months ago, In English,

Hi.

Introduction.

We have a graph, we want to save it (and maybe run DFS on it for example); simple? Yes, but I want to show different methods for this, you can choose any of them depending on problem.

Adjacency matrix.

Simple, if graph is not weighted, save in G[v][u] if there is an edge between v and u or not.

If graph is weighted, if there is an edge between v and u save it’s weight in G[v][u], save  - 1 otherwise (if weight can be negative, use inf instead of  - 1).

Memory usage : .
Overall time complexity : (for running dfs).

Code here

Vectors.

It's the most common method for saving graph. For each vertex keep a vector of it’s edges, now for each edge just save it in related vectors. Use pair (or another struct) for saving weighted graph. It works similar for directed graph.

Memory usage : (Note that vector uses more memory than you need).
Overall time complexity : (for running dfs).

Code here
Dynamic array.

This method is rarely used. For each vertex count edges connected to it at first, then allocate a enough space for each vertex to save it’s adjacency-list. Note that this method is Offline.

Memory usage : .
Overall time complexity : (for running dfs).

Code here

Linked list.

This method is used when id of edges are important and works only for directed graphs (or if you can convert the undirected graph to directed). We use array for saving edges. Then, for each vertex we keep index of last added edge and for each edge consider it’s from Fromi to Toi, we keep the index of last edge before this edge that Fromindex = Fromi and name it prvi. Then we can use headv and prv array to find adjacency-list.

Memory usage : .
Overall time complexity : (for running dfs).

Code here

Application: Solving flow network problems. Wikipedia: Maximum flow problem.

Naive implementation
Dinic’s blocking flow algorithm

Usage of this method in above codes are where we use cap[e] -= x, cap[e ^ 1] += x. Because of special adding edge method, it is guaranteed that if edge e is from v to u, is its pair and it’s from u to v.

Related submission : 23115110.

Index keeping.

This method is used when index of edges is important (i.e. in queries we need to change something about edges, i.e. change weight, delete edge). Keep index of edges related to each vertex in its vector, and for each edge keep the vertices connecting with this edge (see code below).

Note that index keeping is possible and maybe easier using map (or unordered_map), but it is faster using this method. Also note that dynamic array method could be used here as well.

Memory usage : (Note that vector uses more memory than you need).
Overall time complexity : (for running dfs).

Code here

Example: Consider you need to change weight of edges online, this is simple using index keeping method, but it isn’t possible with other methods (at least it will be harder). Related submission : 20776118.

Application: Finding Euler tour. Wikipedia : Eulerian path. Blocking edges is easier with this method, that’s what we need in Euler tour finding. Related submissions : 21395473, 21636007.

Comparison.

I used Codeforces polygon for comparing and generated 16 random graphs:

  • #1 to #4 : n = 106, m = 106

  • #5 to #4 : n = 105, m = 106

  • #9 to #8 : n = 104, m = 106

  • #13 to #16 : n = 103, m = 106

# dynamic array linked list vector
1 OK 1637 / 44 OK 1123 / 29 OK 1606 / 39
2 OK 1637 / 44 OK 1247 / 29 OK 1559 / 39
3 OK 1590 / 44 OK 1107 / 29 OK 1560 / 39
4 OK 1669 / 44 OK 1169 / 29 OK 1622 / 39
5 OK 1060 / 29 OK 1060 / 31 OK 1169 / 36
6 OK 1106 / 29 OK 1045 / 31 OK 1123 / 36
7 OK 1060 / 30 OK 1029 / 31 OK 1122 / 36
8 OK 1091 / 29 OK 1044 / 31 OK 1169 / 36
9 OK 1029 / 26 OK 1013 / 29 OK 1044 / 29
10 OK 982 / 26 OK 998 / 29 OK 1013 / 29
11 OK 1013 / 26 OK 982 / 29 OK 1201 / 29
12 OK 1123 / 26 OK 1029 / 29 OK 1029 / 29
13 OK 998 / 26 OK 951 / 29 OK 982 / 27
14 OK 935 / 26 OK 982 / 29 OK 950 / 27
15 OK 951 / 26 OK 951 / 29 OK 982 / 27
16 OK 966 / 26 OK 1060 / 29 OK 1013 / 27
Σ 16 16 16
max. 1669ms / 44MB 1247ms / 31MB 1622ms / 39MB

It's really strange for me that vector uses less memory than dynamic array while testing, can someone explain the reason? Here is the link to codes and generator.

UPD: I found the answer, I missed that in dynamic array method we are saving edges in pairs, and keeping the sz array in addition (thanks to ATofighi).

As usual I’d like to finish the post with a poem:

بی سبب هرگز نمی‌گردد کسی یار کسی
یار بسیار است تا گرم است بازار کسی

By: Parto biza’ee arani.

Translation: No one becomes friend with another one without reason. People are like shops, a shop is full of customers while it has goods.

P.S. Kindly write in comments (or use private chat in case of necessary) if something is wrong.

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By Arpa, history, 10 months ago, In English,

Hi!

I've recently found Spoj ToolKit. It's features:

  • Random number, array, string, tree, graph generation.
  • A big database of correct solutions to check if your output is correct for custom input.

Let's help this website by adding more correct solutions to its database.

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By Arpa, history, 12 months ago, In English,

Hello again, It’s Arpa as usual :P. Hope you enjoyed from the Gym, and it’s here is the editorial.

Problem packages are available here. Solutions are available here separately.

Preparation details:

The problems authored by me when the main contest was authoring. Here is a table, showing the percentage of expected accepts (in my opinion, before the contest) and the number of accepts.

A BCD
Expected 100% 30%70%10%
Accepted 27 080

Difference from main problems

A : n is bigger, binary_pow will not work. You need to calculate in a faster way.

B : n and also numbers are much bigger, simple xor will not work.

C : n is bigger, simple LCM will not work because the answer would become really large.

D : Number of alphabets are 26 instead of 22, so it isn’t possible to allocate an array with size 2z (z = 26).

Hints

A : Write the last digit of 1378n for several small values. Calculate in a fast way.

B : Note that if then . Use trie.

C : If the answer exists, it depends on the lengths of cycles in the functional graph. Factorize numbers and calculate LCM.

D : Keep a mask for each vertex, i-th bit of maskv is true if the number of edges in the path from the root to v such that letter i is written on them is odd. Now if number of bits in is 0 or 1, path between v and u is Dokhtar-kosh. Give an id to each mask.

Details

First accepts: A : retired_coder, C : wrinx.

Funniest code : hrzuser ’s code : 23229169. He was managing to get accepted in problem C, with a solution :O.

Solutions

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I’d like to finish the editorial with the below poem by Saadi Shirazi:


مقدار یار همنفس چون من نداند هیچ کس ماهی که بر خشک اوفتد قیمت بداند آب را

Translation: No one knows the value of good friend as I know, fish will know the value of water when it falls on the beach.

Goodbye ; )

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By Arpa, history, 12 months ago, In English,

Hi !

As I had promised before, hard version of round #383 will be held on Thursday, 22nd December 15:05 UTC. There will be 4 problems, each problem is a harder version of some problem in round #383. The contest is prepared by me. Please don’t copy your codes from previous contest.

I’d like to thank myself as usual, then Mike MikeMirzayanov Mirzayanov for the great Codeforces and Polygon platforms.

P.S. Prepare your fast input / output streams :P

UPD. Note that the Gym is just for training.

TIME CHANGED. Really sorry, but because "Samcode 2016 round #2" will be held tonight (13 to 16 UTC), the gym will be held on Thursday.

UPD. Registration is now open. Note that problems are not sorted according the difficulty.

UPD. Contest is over. Congratulations to winners:

  1. KingArthur

  2. SlavaSSU

  3. team CQBZ找虐组 : JeremyGuo, liujunhao, yuanxinyu402

  4. vipsharmavip

  5. hrzuser :-|

Editorial.

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By Arpa, history, 12 months ago, In English,

(This blog has been restored as Errichto and some other users had wanted.)

Hi !

Here is some implementation for solving RMQ (Tarjan’s algorithm) (Range Maximum / Minimum Query).

It’s very simple to implement and it’s time complexity is O((n + qa(n)), a() stands for Akerman inverse function used in DSU.

Problem : Given array a of n integers, and q queries, for each query print the maximum value in range [L, R].

Solution: We need a array of vectors, called assigned. assigned[r] contains queries that their R is r. When getting queries, push each query in assigned[R]. We need a dsu, first pari is i. We need a stack, named st.

For i from 0 to n, do:
	While st is not empty and a[st.top] <= a[i]
		Set i parent of st.top in dsu and pop this element from st.
	Push i to st
	For each query assigned to i
		Answer of this query is a[root of L of this query in DSU].
Code here.

Note that in above code I used path-compression technique for dsu only, size-comparing technique can be used too (but it has lower performance).

It’s obviously true, because each time for any j ≤ i, a[root(j)] is the greatest value in range [j, i].

Performance test

This method (known as Arpas trick)
Vector + Binary search
Sparse table
O(n) method
generator

Here is the result:

Method\Time(milliseconds)Strictly increasing arrayStrictly decreasing arrayRandom
This method (known as Arpa's trick) 294328902946
Sparse table 361235953807
Vector + Binary search 310161303153
O(n) method 378839203610

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By Arpa, history, 12 months ago, In English,

Hello again, and hope you have been Joon-Joon of the round :P

I’m preparing harder version of some of the problems (Div.2 A, B, and Div.1 A, D) and I’ll put them on some gym, so if you are interested in harder version of problems please wait for about one week.

You can see and download the problem archives (pretests, tests, statements, validator, checker, solutions) here. You can see solutions in solutions folder, note that there are several codes there, there is a descriptor for each code (.desc) that shows the verdict of that code.

Preparation details:

On 25 May Sa1378 came up with problem Div.1 D and other problems authored inchmeal.

9/25/16: Proposal sent to Gleb.

11/12/16: Answer from Gleb: Your proposal has been redirected to Nikolay KAN.

11/15/16: Nikolay has replied, and commented about problems, saying some problems are easy, some are hard, some are ok.

I changed some of the problems, change the constraints for some others and we talked about solutions through email.

11/17/16: We switched to Telegram.

Creating tests, writing generators, writing statements, etc started in the Polygon.

12/06/16: Round #383 hold.

gKseni told me how you want to get your money and I had no idea.

gKseni told me that it is possible to transfer my money and finally May 4, 2017, I received my money through Okpay.

I have another problem set to prepare another div.1 + div.2 round, but I haven’t enough time now :(.

Paintings in problems were suggested by me, painted by Sa1378 and edited by me. Problem stories and this editorial were suggested and written by me. KAN helped us preparing the round very much, we are thankful to him. This table for each person and for each problem shows the number of the committed changes (in polygon) he has made in preparing the problem (it is good for showing how much someone was involved in preparing).

I used Google Docs for writing everything.

User\Problem Div.2 A Div.2 BDiv.1 ADiv.1 BDiv.1 CDiv.1 DDiv.1 ETotal
Me891616101937114
Sa1378321070114
KAN and testers 3359771361

Here is another table, showing the number of expected accepts (in my opinion, before the contest) and the number of accepts (after system testing).

Div.2 A Div.2 BDiv.2 CDiv.2 DDiv.2 EDiv.1 ADiv.1 BDiv.1 CDiv.1 DDiv.1 E
Expected 6000450020005001506004003005010
Accepted 3966172311316841047947450151

Hints

Div.2 A: Write the last digit of 1378n for several small values.

Div.2 B : Note that if then .

Div.1 A: If the answer exists, it depends on the lengths of cycles in the functional graph.

Div.1 B: It’s similar to a simple knapsack problem, think on O(n·W) solution using dynamic programming.

Div.1 C: Build a graph and put edges between each 2 * i, 2 * i + 1 and each BF and GF.

Div.1 D: Keep a mask for each vertex, i-th bit of maskv is true if the number of edges in the path from root to v such that letter i is written on them is odd. Now if number of bits in is 0 or 1, path between v and u is Dokhtar-kosh.

Div.1 E: Sort all of the options, then the problem becomes easier, solve the new problem with sqrt decomposition.

Details

Div.2 A

Idea, authoring, solution by Sa1378, preparation by Sa1378 and me.

My and Sa1378 ’s birth year in Solar Hijri calendar is 1378.

Div.2 B

Idea, authoring, solution by Sa1378, preparation by Sa1378 and me.

Div.1 A

Idea, authoring, solution, preparation by me.

Attractive boys/girls are called Joon-Joon in Persian. Owf is a sound used when we (Persian) are interested in something, especially when we see something attractive, such as our crush :P

Div.1 B:

Idea, authoring, solution, preparation by me.

The problem authored by me 2 days before the contest :D (#FastAsFerrari). Attractive girls are called (some word similar to) “Hos” in Persian. It’s a good place to thank Amsen, whose name (Hir) gave me this idea (to use word “Hos” instead of “attractive girl”).

Div.1 C :

Idea, authoring, solution by Sa1378, preparation by Sa1378 and me.

“Kooft” is something make people die. “Zahre-mar” meaning is “Venom of Snake”.

Div.1 D:

Idea by Sa1378, authoring, solution, preparation by me.

“Dokhtar-kosh” is an adjective, used when something is very very attractive.

Div.1 E:

Idea, authoring, solution, preparation by me.

Solutions

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I’d like to finish the editorial with the below poem by Hafez:


از صدای سخن عشق ندیدم خوش‌تر یادگاری که در این گنبد دوار بماند

Translation: I have never seen anything that sounds better than love, it’s the relic which will remain in the universe.

Good luck and see you soon in “Round #383 hard version” ;)

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By Arpa, history, 12 months ago, In English,

Hi!

I'm honored to invite you to Codeforces Round #383, it will be held on 6nd December 14:35 UTC. There will be 5 problems for each division as usual. The contest was prepared by AmirReza Arpa PoorAkhavan and Mehrdad Sa1378 Saberi. It's our first official contest at CodeForces.

The contest stories will be about Arpa and Mehrdad and some events happen with them in Arpa's land, in addition you will get some information about Arpa's land and girls living there (Owf (t = 1)).

I'd like to thank myself (:P) and Mehrdad at first, then Nikolay KAN Kalinin for helping me in preparing problems and Mike MikeMirzayanov Mirzayanov for the great Codeforces and Polygon platforms.

The scoring distribution will be announced later.

Answer for one of your common questions : -Yes, It is rated.

UPD. GL & HF. Hope you came up with Dokhtar-kosh solutions for our Dokhtar-kosh problems.

Urgent information from MikeMirzayanov: due to hardware issues, the round is moved to Tuesday 6th December, 14:35 UTC. We are very sorry this happened. More information is available in this post.

UPD. Scoring distribution: Div.1 : 500-1000-1250-2000-2500, Div.2 : 500-1000-1500-2000-2250.

UPD. Contest is over, hope you have been Joon-Joon of the round :P

Congratulations to winners:

Div.1:

1 . moejy0viiiiiv

2 . mnbvmar

3 . data_h and nuip (WoW :O)

5 . Phronesis

Sepcial congratulations to anta who solved Div.1 E.

Div.2:

1 . gilcu3

2 . toHisDream

3 . Fery

4 . 768092

5 . orz_liuwei

Editorial is ready.

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By Arpa, history, 14 months ago, In English,

Hi !

After a long delay I want to publish INOI (Iran National Olympiad of Informatics) results :

Gold medal :

  1. Seyed Mohammad Hossein Nematollahi (Narenji58).

  2. amir azarmehr (AmirAz).

  3. Mohammad Saneian (SaSa).

  4. HamidReza Hedayati (NeoBit).

  5. Ali Ahmadi (Kuzey).

  6. Ali Shafiei (Ali.Sh).

  7. Iman Gholami (IMAN_GH).

  8. Majid Garoosi (MaGaroo).

Silver medal (Only Top5 and last) :

  1. Yara Kamkar (yarak).

  2. Seyed Hossein Moosavi (mr_agha_seyed).

  3. Shayan CheshmJahan (Shayan_Jahan).

  4. Soroosh Taslimi (Information_schema) (Thanks to Navick).

  5. Mehrdad Saberi (Sa1378).

...

16 . Keyvan Rezaei (Navick) (I added him because of AmSen's Comment :D)

I want to congratulate my best friends, Shayan CheshmJahan, Mehrdad Saberi, Seyed Mohammad Hossein Nematollahi and specially Ali Shafiei, and my other friends. And I wish the bests for my friend Yara Kamkar in next steps of life.

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By Arpa, history, 20 months ago, In English,

Changes are available in history section.

Hi!

Most of the people know about dsu but what is the "dsu on tree"?

In Iran, we call this technique "Guni" (the word means "sack" in English), instead of "dsu on tree".

I will explain it and post ends with several problems in CF that can be solved by this technique.

What is the dsu on tree?

With dsu on tree we can answer queries of this type:

How many vertices in the subtree of vertex v has some property in time (for all of the queries)?

For example:

Given a tree, every vertex has color. Query is how many vertices in subtree of vertex v are colored with color c?

Let's see how we can solve this problem and similar problems.

First, we have to calculate the size of the subtree of every vertice. It can be done with simple dfs:

int sz[maxn];
void getsz(int v, int p){
    sz[v] = 1;  // every vertex has itself in its subtree
    for(auto u : g[v])
        if(u != p){
            getsz(u, v);
            sz[v] += sz[u]; // add size of child u to its parent(v)
        }
}

Now we have the size of the subtree of vertex v in sz[v].

The naive method for solving that problem is this code(that works in O(N ^ 2) time)

int cnt[maxn];
void add(int v, int p, int x){
    cnt[ col[v] ] += x;
    for(auto u: g[v])
        if(u != p)
            add(u, v, x)
}
void dfs(int v, int p){
    add(v, p, 1);
    //now cnt[c] is the number of vertices in subtree of vertex v that has color c. You can answer the queries easily.
    add(v, p, -1);
    for(auto u : g[v])
        if(u != p)
            dfs(u, v);
}

Now, how to improve it? There are several styles of coding for this technique.

1. easy to code but .

map<int, int> *cnt[maxn];
void dfs(int v, int p){
    int mx = -1, bigChild = -1;
    for(auto u : g[v])
       if(u != p){
           dfs(u, v);
           if(sz[u] > mx)
               mx = sz[u], bigChild = u;
       }
    if(bigChild != -1)
        cnt[v] = cnt[bigChild];
    else
        cnt[v] = new map<int, int> ();
    (*cnt[v])[ col[v] ] ++;
    for(auto u : g[v])
       if(u != p && u != bigChild){
           for(auto x : *cnt[u])
               (*cnt[v])[x.first] += x.second;
       }
    //now (*cnt[v])[c] is the number of vertices in subtree of vertex v that has color c. You can answer the queries easily.

}

2. easy to code and .

vector<int> *vec[maxn];
int cnt[maxn];
void dfs(int v, int p, bool keep){
    int mx = -1, bigChild = -1;
    for(auto u : g[v])
       if(u != p && sz[u] > mx)
           mx = sz[u], bigChild = u;
    for(auto u : g[v])
       if(u != p && u != bigChild)
           dfs(u, v, 0);
    if(bigChild != -1)
        dfs(bigChild, v, 1), vec[v] = vec[bigChild];
    else
        vec[v] = new vector<int> ();
    vec[v]->push_back(v);
    cnt[ col[v] ]++;
    for(auto u : g[v])
       if(u != p && u != bigChild)
           for(auto x : *vec[u]){
               cnt[ col[x] ]++;
               vec[v] -> push_back(x);
           }
    //now (*cnt[v])[c] is the number of vertices in subtree of vertex v that has color c. You can answer the queries easily.
    // note that in this step *vec[v] contains all of the subtree of vertex v.
    if(keep == 0)
        for(auto u : *vec[v])
            cnt[ col[u] ]--;
}

3. heavy-light decomposition style .

int cnt[maxn];
bool big[maxn];
void add(int v, int p, int x){
    cnt[ col[v] ] += x;
    for(auto u: g[v])
        if(u != p && !big[u])
            add(u, v, x)
}
void dfs(int v, int p, bool keep){
    int mx = -1, bigChild = -1;
    for(auto u : g[v])
       if(u != p && sz[u] > mx)
          mx = sz[u], bigChild = u;
    for(auto u : g[v])
        if(u != p && u != bigChild)
            dfs(u, v, 0);  // run a dfs on small childs and clear them from cnt
    if(bigChild != -1)
        dfs(bigChild, v, 1), big[bigChild] = 1;  // bigChild marked as big and not cleared from cnt
    add(v, p, 1);
    //now cnt[c] is the number of vertices in subtree of vertex v that has color c. You can answer the queries easily.
    if(bigChild != -1)
        big[bigChild] = 0;
    if(keep == 0)
        add(v, p, -1);
}

4. My invented style .

This implementation for "Dsu on tree" technique is new and invented by me. This implementation is easier to code than others. Let st[v] dfs starting time of vertex v, ft[v] be it's finishing time and ver[time] is the vertex which it's starting time is equal to time.

int cnt[maxn];
void dfs(int v, int p, bool keep){
    int mx = -1, bigChild = -1;
    for(auto u : g[v])
       if(u != p && sz[u] > mx)
          mx = sz[u], bigChild = u;
    for(auto u : g[v])
        if(u != p && u != bigChild)
            dfs(u, v, 0);  // run a dfs on small childs and clear them from cnt
    if(bigChild != -1)
        dfs(bigChild, v, 1);  // bigChild marked as big and not cleared from cnt
    for(auto u : g[v])
	if(u != p && u != bigChild)
	    for(int p = st[u]; p < ft[u]; p++)
		cnt[ col[ ver[p] ] ]++;
    cnt[ col[v] ]++;
    //now cnt[c] is the number of vertices in subtree of vertex v that has color c. You can answer the queries easily.
    if(keep == 0)
        for(int p = st[v]; p < ft[v]; p++)
	    cnt[ col[ ver[p] ] ]--;
}

But why it is ? You know that why dsu has time (for q queries); the code uses the same method. Merge smaller to greater.

If you have heard heavy-light decomposition you will see that function add will go light edges only, because of this, code works in time.

Any problems of this type can be solved with same dfs function and just differs in add function.

Hmmm, this is what you want, problems that can be solved with this technique:

(List is sorted by difficulty and my code for each problem is given, my codes has heavy-light style)

600E - Lomsat gelral : heavy-light decomposition style : 14607801, easy style : 14554536. I think this is the easiest problem of this technique in CF and it's good to start coding with this problem.

570D - Tree Requests : 17961189 Thanks to Sorasorasora; this problem is also good for start coding.

Sgu507 (SGU is unavailable, read the problem statements here) This problem is also good for the start.

HackerEarth, The Grass Type This problem is also good for start (See bhishma's comment below).

246E - Blood Cousins Return : 15409328

208E - Blood Cousins : 16897324

IOI 2011, Race (See SaSaSaS's comment below).

291E - Tree-String Problem : See bhargav104's comment below.

343D - Water Tree : 15063078 Note that problem is not easy and my code doesn't use this technique (dsu on tree), but AmirAz 's solution to this problem uses this technique : 14904379.

375D - Tree and Queries : 15449102 Again note that problem is not easy :)).

716E - Digit Tree : 20776957 A hard problem. Also can be solved with centroid decomposition.

741D - Arpa’s letter-marked tree and Mehrdad’s Dokhtar-kosh paths : 22796438 A hard problem. You must be very familiar with Dsu on tree to solve it.

For Persian users, there is another problem in Shaazzz contest round #4 (season 2016-2017) problem 3 that is a very hard problem with this technique.

If you have another problem with this tag, give me to complete the list :)).

And after all, special thanks from amd who taught me this technique.

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By Arpa, 21 month(s) ago, In English,

Hi !

Today while solving 356D - Монеты и мешочки I needed a function for bitset in order see what is the first set bit.I asked M.Mahdi and he told me about bs._Find_first(). for example:

bitset<17>BS;
BS[1] = BS[7] = 1;
cout<<BS._Find_first()<<endl; // prints 1

After more research , we found bs._Find_next(idx). This function returns first set bit after index idx.for example:

bitset<17>BS;
BS[1] = BS[7] = 1;
cout<<BS._Find_next(1)<<','<<BS._Find_next(3)<<endl; // prints 7,7

So this code will print all of the set bits of BS:

for(int i=BS._Find_first();i< BS.size();i = BS._Find_next(i))
    cout<<i<<endl;

Note that there isn't any set bit after idx, BS._Find_next(idx) will return BS.size(); same as calling BS._Find_first() when bitset is clear;

UPD One question, bitset is 32 or 64 times faster than bool array?

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By Arpa, 23 months ago, In English,

Hi!

I thought that unordered_set is faster than set, because set complexity is logarithmic in size and unordered_set complexity is constant time in average.(source: cplusplus.com)

But today I saw this:

TLE(>2000 MS) with unordered_set: 15494816

Accepted(155 MS) with set: 15494846

Also my other solution(witch I submitted during the contest) with unordered_map hacked :'(

Now my question is : Is unordered_set faster than set?

UPD: I have accepted(15497282).unordered_set time was better than set(15494846) : 93<155.

Only with adding this line:s.max_load_factor(0.25);s.reserve(500);.

UPD2: it seems that it is better to use a power of 2 in reserve(i.e. s.reserve(512)).(Thanks from keyvankhademi)

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By Arpa, history, 23 months ago, In English,

Hi!

One of the C++ programmers problems is to work with integers greater than 2^64-1 (we can save 0to 2^64-1 in unsigned long long int). So I want to share the best Bignum implementation I have ever seen (Link) with CodeForces Community.

Its specifications are as follows:

  • Supported operations: + , -, / , * , % , ^(pow) , gcd , lcm , abs.

  • It is able to work with Standard Input/Output streams.

  • It can cast data to long long, string.

  • It uses fast multiplication.

source.(but I have edited that and added pow and size().)

UPD1 (September 2016): Bug in void operator=(long long v) is now fixed. Thanks to AmSen.

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By Arpa, 23 months ago, In English,

Hi!

I am in virtual contest and I'm seeing this in my friends standings.

What is the problem?

P.S. top10 in contest:

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By Arpa, 2 years ago, In English,

Hi!

After my previous post about unordered_map now I want to explain hash functions.

std::hash.

C++ STL has one hash function in library <functional>. You can use it for this data types:

template<> struct hash<bool>;
template<> struct hash<char>;
template<> struct hash<signed char>;
template<> struct hash<unsigned char>;
template<> struct hash<char16_t>;
template<> struct hash<char32_t>;
template<> struct hash<wchar_t>;
template<> struct hash<short>;
template<> struct hash<unsigned short>;
template<> struct hash<int>;
template<> struct hash<unsigned int>;
template<> struct hash<long>;
template<> struct hash<long long>;
template<> struct hash<unsigned long>;
template<> struct hash<unsigned long long>;
template<> struct hash<float>;
template<> struct hash<double>;
template<> struct hash<long double>;

For example:

hash<int>hi;
int x=69;
cout<<hf(x)<<endl;
hash<long double>hld;
long double y=69.6969696969;
cout<<hld(y)<<endl;
hash<vector<int> >hv;
vector<int>v({69,69,69,69});//v→ 69,69,69,69
cout<<hv(v)<<endl;
Hand-made hash functions.

There are many ways to create hash function for your struct. For example:

struct reval{
  vector<int>v;
  int n;
  string s;
  size_t hash(){//size_t is alias of unsigned int
    hash<string>hs;
    hash<long long>hll;
    hash<int>hi;
    hash<vector>hv;
    long long ans=hs(s);
    ans<<=32;
    ans|=hi(n);
    ans=hll(ans);
    ans<<=32;
    ans|=hv(v);
    ans=hll(ans);
    return ans;
  }
}

A simple example:(be careful;it is only a sample and it isn't good hash function of course)

struct reval{
  vector<int>v;
  int n;
  string s;
  size_t hash(){//size_t is alias of unsigned int
    hash<string>hs;
    hash<int>hi;
    hash<vector>hv;
    return hs(s)+hv(v)+hi(n);
  }
}

A trick:

struct S{
  string first_name;
  string last_name;
};
namespace std{
    template<>struct hash<S>{
        size_t operator()(S const& s) const{
            size_t h1=hash<string>()(s.first_name);
            size_t h1=hash<string>()(s.last_name);
            return hash<long long>()( (h1<<32)^h2 );
        }
    };
}
int main(){
    S s;
    s.first_name="MohammadSina";
    s.last_name="PakSeresht";
    cout <<"hash(s) = "<<hash<S>()(s)<<endl;
}

Note1: You can use hash<type>()(variable), like this:

int x=69;
cout<<hash<int>()(x)<<endl;

Instead of:

int x=69;
hash<int>hasher;
cout<<hasher(x)<<endl;

Note2: please be careful about combining two hashes. There are two good ways:

1-x*P+y mod M(when P is prime number (like 43) and M is less than 2^32 and not a power of 2 (like 10^9+7)).

2-hash<long long>()((x<<32)^y) or hash<long long>()(x*1000000007LL+y).

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By Arpa, history, 2 years ago, In English,

UPD: Tricks to make unordered_map faster added.

Hi!

What is unordred_map?

It is a data structure like map but it is more than 4 times faster than map.you can use it in C++11 with including #include<unordered_map>.for example:

#include<unordered_map>
using namespace std;
int main(){
  unordered_map<int,int>mp;
  mp[5]=12;
  mp[4]=14;
  cout<<mp[5]<<' '<<mp[4]<<endl;//prints: 12 14
}

Lets explain it more.

How it works?

Focus on unordered_set for simplify.You can imagine that it has vector of vector like vector<vector<type> > mp. Every time you insert value V in that, it calculate its hash(I will explain how it works), let hash(V)=K; it inserts V into mp[K] (formally it calls mp[K].push_back(V)).When you call mp.count(V) it searchs for V in mp[K].

map VS unordered_map (and set VS unordered_set)

1-unordered_map is more than 4 times faster

Focus on problem 527E - Data Center Drama, it seems that it is good to use unordered_map instead of map.

My submission with map: 14269000 Time:484 MS.

My submission with unordered_map: 14269381 Time:358 MS.

Another good sample to show how fast is unordered_map is problem 178C3 - Smart Beaver and Resolving Collisions:

My submission with map: 15781810 Time limit exceeded on test 36 .

My submission with unordered_map: 15774685 Accepted (Time:966 MS).

2-unordered_set (and unordered_map) is not sorted

Please note that set (and map) is sorted, for example *(s.begin()) is always smallest number in the set; but in unordered_set it isn't. similarly there isn't lower_bound and upper_bound in unordered_set (and unordered_map similarly).

Creating hash function for structs

Now, one other problem remains, you can try to compile this code:

unordered_map<pair<int,int>,int>mp;

You will get Compilation Error! Why? See this page for unordered_map supported types. For unsupported types, you have to create your own hash function for use. For example lets see how we can create a hash function for pair<int,int>.

As you know any int value is between -2^31+1 to 2^31-1.so if we create function that for every pair<int,int> returns distinct value in type size_t(alias of unsigned int), it will be done. It is pretty easy: x.first^(x.second<<32) is good. but be careful about overflow ;for having good hash function we use hash<long long>.The code is looks like this:

struct HASH{
  size_t operator()(const pair<int,int>&x)const{
    return hash<long long>()(((long long)x.first)^(((long long)x.second)<<32));
  }
};
unordered_map<pair<int,int>,int,HASH>mp;

Now you have a unordered_map of pair<int,int> (it isn't problem what is second member, in example it was int).Creating hash function for other structs is same.

How to test unordered_map is faster than map?

You can test that for N=10^6, unordered_set(unordered_map) is more than 4 times faster than set(map) ;with this code:(compile code with command: g++ file.cpp -std=c++11 -O2)

#include <bits/stdc++.h>
using namespace std;
unordered_set<int>s;//replace it with set for test.
int main(){
  auto T=clock();
  s.reserve(32768); //updated !
  s.max_load_factor(0.25); //updated !
  for(int i=0;i<1000000;i++)
    s.insert(i);
  cout<<double(clock()-T)/CLOCKS_PER_SEC<<'\n';
}

Note1: Let your hash function H(V), it is better that H(V) returns distinct value for every distinct V, it makes unordered_map faster; but if you can't do that it doesn't problem. The only problem is that unordered_map becomes slower(however it is better than map).

Note2: Please be careful about your hash function time complexly! it is fun to use this hash function:

struct HASH{
  size_t operator()(const pair<int,int>&x)const{
    size_t ans=0;
    for(int i=0;i<x.first;i++)
      ans+=x.second;
    return ans;
  }
};

UPD I will explain hash<type> in my next post.

UPD It seems that sometimes unordered_map becames so slow.but it can improve with this two lines of code:

unordered_map<int,int>mp;
mp.reserve(1024);
mp.max_load_factor(0.25);

With this two lines unordered_map become about 10 times faster. You can replace 1024 with another suitable power of two.(it depends on number of insert s you will do).

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By Arpa, history, 2 years ago, In English,

Hi!

COCI (CROATIAN OPEN COMPETITION IN INFORMATICS) will be held saturday.

Let's discuss about problems after contest.

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By Arpa, history, 2 years ago, In English,

Hello

it will be good for all programmers :)

https://sercantutar.github.io/infint/

Geek trick: you can paste all InfInt.h to top your code for sending it for CF or another programming sites :D

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