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Автор SecondThread, история, 4 года назад, По-английски

Algorithms Thread Episode 8: Tree Basics

Episode 8 of Algorithms Thread comes out in <90 minutes! This one is a bit more beginner-friendly and covers the following ideas:

  • Graph/Tree Diameters
  • Binary Lifting
  • Tree Flattening with Euler tours

Also, to make sure you have actually learned that stuff, I made a custom Gym set on CodeForces that will last two weeks that hopefully is really good practice for making sure you have learned this stuff. Here is a link to the gym set; it will be available 45 minutes after the video comes out so that people have time to watch the video before starting the set, if they are interested in penalty points. All of the problems in the gym are original to this set (in their flavortext at least, some are simple enough that I'm sure they have appeared in other contests before).

The new gym integration was heavily inspired by Errichto's Matrix Expo set format. Let me know whether it's helpful. I think it might be, but also it's a pretty big time commitment to make it, so whether I keep doing them depends on how helpful they are to people.

If you have any questions or suggestions, feel free to leave them below. I hope you enjoy the problem statements, and I'll leave you all with this:


Solutions

Update:

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4 года назад, # |
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Really appreciate your effort for helping beginners to learn new concepts and improve!

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4 года назад, # |
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Looking forward to it!

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4 года назад, # |
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    4 года назад, # ^ |
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    Hey, I understand that you want to make a point, but don't you think tagging them is unnecessary? I believe noone likes to be tagged just to see random people being orzed.

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      4 года назад, # ^ |
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      Guys, please listen to Ari, don't tag anyone unnecessarily.

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        4 года назад, # ^ |
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        Stop tagging Ari, She doesn't like it

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          4 года назад, # ^ |
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          Hey, I understand that you guys want to make a point, but don't you think tagging Ari is unnecessary? I believe no one likes to be tagged just to see themselves being joked about by random people.

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      4 года назад, # ^ |
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      I fully support condemnation of this practice. Looking at a notification just to see myself being tagged unnecessarily by random people is sure to be tiring.

      In this case, however, I do not think it is a harm. It is a blog written by the "orzed" person, and he (assumingly) reads and enjoys comments here.

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      4 года назад, # ^ |
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      No, no, why are you saying this? I'm pretty sure that neugis tagged me for a really good reason.

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      4 года назад, # ^ |
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      Or SecondThreaded

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      4 года назад, # ^ |
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      For the record, I like being pinged so that I know when people are talking about me. At least, I certainly don't mind.

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4 года назад, # |
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Wait..... How come this is on the frontpage of CF?

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4 года назад, # |
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Errichto and Thread are changing the learning game istg.

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4 года назад, # |
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At this point, since everything's already prepared for it, they might as well add your stuff to EDU

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4 года назад, # |
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Spreading CP knowledge!
Screenshot-from-2020-08-17-01-07-07
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4 года назад, # |
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Thanks, we will try our best to learn new and interesting concepts.

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4 года назад, # |
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Loved the new format ! This helps gain much more confidence in the covered than just watching a tutorial on some topic and not getting custom problems on that topic.

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4 года назад, # |
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Thanks, that’s exactly what I need.

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4 года назад, # |
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The gym problems are very nice for reaffirming one's understanding of tree basics!

For problem D, what was the offline solution you had in mind? The text at the top implies that there exists an offline solution that's simpler than the online solution, but the only thing I could think of is some wack dfs + segtree solution where you update the answer for a query whenever you reach one of its endpoints. That seems way too complicated to warrant explicitly disallowing offline solutions lol.

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    4 года назад, # ^ |
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    Yeah it's definitely harder to do offline than online. I was also considering at a small-to-big + dfs with a treeset/segtree which I'm pretty sure is possible and can be done in n*log^2(n) without too much difficulty. The main thing is that this might show up as a sub-problem that you have to answer online, so it's good to be able to do that, rather than finding some complex way that ACs but doesn't actually use the new topics at all.

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4 года назад, # |
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I really love your content

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4 года назад, # |
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Your screencast(s) are really helpful.

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4 года назад, # |
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Please give any hints for second problem.

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4 года назад, # |
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Dear SecondThread,

Thank you for all the hard work you put in daily! Know that it is recognized and greatly appreciated, It’s an honor for our codeforces community to have someone like you.

May you reach every height of success!

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4 года назад, # |
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Hey nice video,

In the tree shown at 36:11, how can I find the sum over the path starting at the node with input time 3 and the node with input time 13? I can think of a way to do it but it requires calculating LCA of the endpoints, is there a way to do it without LCA?

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    4 года назад, # ^ |
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    Thanks! Correct, if you need to find some path aggregate in which one node isn’t an ancestor of another, you need to split it into two paths using LCA. (Unless there is some other more clever way that I’m not aware of)

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4 года назад, # |
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Can Someone Explain the Samples of Problem F Please

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    4 года назад, # ^ |
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    Yeah, maybe part of it isn't super clear: for each query where x = 0, you can consider pairing things however you want just before for that query. These pairings aren't permanent, they are just considerations for that query. The way of making the pairings is to do minimize the sum of distances between every pair of paired seeds/pots.

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4 года назад, # |
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expecting more and more from all of you

thanks SecondThread

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4 года назад, # |
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The gym-set along with the usual videos format is indeed really nice. Thank you for your hard work.

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4 года назад, # |
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I've just failed my math exam because of secondthread, he says that pi = 3 in 102694A - Circumference of a Tree, so I used pi = 3 to calculate the volume of a sphere and I got 0 marks on that problem. Guys, don't trust secondthread.

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4 года назад, # |
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A bit off topic, but can we expect a video on round 1 problems of hackercup, SecondThread ?

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    4 года назад, # ^ |
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    He said he won't be taking part in hacker cup in one of his streams.

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      4 года назад, # ^ |
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      I think...he said something along the lines that he wasn't sure if would be allowed to....but I might be wrong...

      Also, I found a comment on youtube...where he answered it so I guess that's it for this thread! (which is a cool line that I am hoping will catch on...lol)

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    4 года назад, # ^ |
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    I’m not allowed to compete in hackercup, so probably not from me. I know Errichto has videos of him solving them, and there are some nice solutions on the Hacker cup website, too.

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4 года назад, # |
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Hey guys, I am bit stuck on prblem B, my approach was:

  • start from arbitary node and find the farthest node, call it 'root'
  • from 'root' , do a bfs and find the distance of all other nodes
  • The maximum distance now will be the diameter,now,
  • If distance of some node from 'root' is same as diameter or zero, the answer for this will be diameter+1 , else the answer for this node will be same as diameter

I am getting a wrong answer on test 6. Can someone plz help?

Any general tips to debug will also be appreciated.

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    4 года назад, # ^ |
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    There can be multiple "root". Take care of that also

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      4 года назад, # ^ |
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      I took care for that. I tried DFS from any general node..got all the max distant nodes.. Then ran a dfs from all of them to get their end points...of maxDiam..

      Then simply checked if for i is it in the stored points..I added diam+1 as asnwer..else diam is answer.

      Getting TLE for TC4 https://codeforces.com/gym/102694/submission/90401537

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        4 года назад, # ^ |
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        Make sure you dont linearly check, probably some faster way

        Spoiler
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          4 года назад, # ^ |
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          I didn't get the spoiler hint..It would be great if you could eleborate with some another example which could relate things up.

          As I got that searching linearningly for N nodes would difinately be TLE...

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            4 года назад, # ^ |
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            Find the boundary points. Think of a better way than a linear search.

            Spoiler
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              4 года назад, # ^ |
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              There is little confusion. I am using Set for finding the elements if the element is end point of diameter. The only linear part I am doing is with find function which calculates the dep and diam initially.

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                4 года назад, # ^ |
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                Yeah, that's not the intended solution at all. There shouldn't be any need for a log factor here.

                Spoiler
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    4 года назад, # ^ |
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    What if the tree is like this?

    Tree

    From node 1 you will find 7 (or 8) as the farthest node. And then from 7 (or 8) you will find 1 as the only farthest node. The problem is that your algorithm will consider only 1 and 7 (or 1 and 8) as all the possible nodes which should increase the diameter by 1, but actually all of them will increase it (1, 7, 8). You can try this example yourself for more details.

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      4 года назад, # ^ |
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      Hey Brodicico! Could you(or anyone) please tell me what mistake I am making. (I am continuously getting WA on test 5 and I am unable to figure out what mistake I am making — whether it is the implementation or the logic)

      LOGIC :
      CODE :

      A counter test case with its correct output is highly appreciated.

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        4 года назад, # ^ |
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        Try a case where you have a tree in a chain form like 1-2-3-4-5-6-7-8-9. Then connect node 10 with vertex 5.

        The tree will be :

        1-2-3-4-5-6-7-8-9
                |
                10
        

        For node 10 it isn't correct to output diam+1.

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      4 года назад, # ^ |
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      bro i am getting correct answer for this one but still my code fails no 4th test case

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    4 года назад, # ^ |
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    There's a hint in the video..watch it again....

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4 года назад, # |
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This was a wonderful effort . can anyone give me the link for other episodes.

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4 года назад, # |
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Your screencast are really helpful. Thank You!

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4 года назад, # |
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1006E - Military Problem
Great problem on one of the concepts explained in the video. Highly recommend it!

My solution for reference: 90223865

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4 года назад, # |
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Someone give me a hint on E, please

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4 года назад, # |
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As a beginner, your episode has really taught me a lot about the many application of trees. Thanks a ton for this!

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4 года назад, # |
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Hey can someone recommend a good list of problems which contains different( hopefully all theory) types of problems on graphs and/or trees....It would be much appreciated....

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4 года назад, # |
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can someone help me , I am getting memory limit exceeded in B. https://paste.ubuntu.com/p/6B8kdrvkyh/

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    4 года назад, # ^ |
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    Hey buddy, I saw your code, actually I think you can simplify the structure itself. Also the variable parent can be avoided, vis can be made boolean. Have you checked if the bfs doesn't explodes the queue?

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      4 года назад, # ^ |
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      I made vis boolean , now i getting tle. thanks man ,Changing the structure now. changed endl to "\n" got accepted.

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        4 года назад, # ^ |
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        Didn't know "\n" could change a whole lot of things. Followed some blogs on CF on this. Finally got AC after all the TLEs.

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4 года назад, # |
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SecondThread

Any idea for D. Till now , this is how I am going: ( I guess m is always n-1).

  • we need to find the minimum valued edge between sources and destination of the flow network, (as its a tree <or I am wrong here?> and only one path is present between any two nodes).
  • we can do this by storing values of minimum for 1,2,4,8,16.. jumps above for each node, and find them just like we build the jump array in binary lifting.
  • Now we can move to the lca and keep calculating the minimum along both souces to lca and destination to lca, again using the binary lift technique.

Am I going the right way ? ( Wrong answer on 6)

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    4 года назад, # ^ |
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    I do actually have an idea of how to solve D, yeah. Your general approach seems right, but I’m not going to discuss solution details too in depth for a week or so.

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4 года назад, # |
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Someone give me a hint on F, please

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4 года назад, # |
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My 2-dfs calls in B is receiving TLE, is it expected ?

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4 года назад, # |
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My code for problem B https://ideone.com/dd5gOx with poor English commentary :)

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4 года назад, # |
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Math burned me, yet again!!

SecondThread

For E the products can be as large as 10^500000 ( that is ((10^5)^(10^5)) ) and we have to store approx 10^6 of those.....how can it be done....can I get a hint?

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4 года назад, # |
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The problems are really, really good. Even if they use basic concepts they were not actually so easy (at least for me). I found the fifth problem very cool and challenging if you don't have experience using mathematical functions (I didn't until this problem :D). I give all my respects to the creator of this problemset. Thanks for investing your time in helping others.

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4 года назад, # |
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For question C the code is running on my computer btt it shows runtime error for test case 3

90309168

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    4 года назад, # ^ |
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    since contest is running, nobody can open the link you posted. Also since test case 3 is visible you can directly debug in local environment to find reason for RE.just use gdb/lldb to step through code.

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      4 года назад, # ^ |
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      test case 3 is running in my local env just fine. btt it shows runtime error whwn i submit.

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4 года назад, # |
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I haven't registered yet does that mean i'll have to wait for next 12 days?

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4 года назад, # |
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Image:

Hello, i am currently at problem E and i dont understand how does it is not 1.5 in the second example, at the second line.

Since it is 1-rooted, 1 has value 3, 2 has value 2 and 3 has value 1 , no ??? what did i missed here ??

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    4 года назад, # ^ |
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    In the second sample, none of the nodes are ever updated, so all nodes have a value of 1. (Since the product of 1*1*1... is 1)

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4 года назад, # |
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Could anyone please help me with problem E?How to handle such big numbers?

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4 года назад, # |
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I realy enjoy E, although I've made 17 WA.

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4 года назад, # |
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Can someone tell me why bfs doesn't work in promlem A to calculate the diameter. It is giving me tle.

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    4 года назад, # ^ |
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    It worked for me, maybe because you're using python and the time is a little tight, try reading this blog to speed up the input and the output. If it doesnt work to you, maybe trying with c++ will be a better way, because I see that the few persons that got AC, their execution time is around to 900ms

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4 года назад, # |
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Hi, this is my first time solving Tree based problems on CF. I am having trouble understanding how to give the edges so that I can construct a tree. For example in the first question, the test case is

3

3 2

2 1

How will we construct a tree with 3,2 as edges and which node will they belong to? How will we construct a tree with this? I have seen some other solutions using arrays but i can't understand them.

Leetcode does it like this-

struct TreeNode {
     int val;
     TreeNode *left;
     TreeNode *right;
     TreeNode() : val(0), left(nullptr), right(nullptr) {}
     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};

I have previously solved on LeetCode where we had to just write the function so I am very new to this.

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    4 года назад, # ^ |
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    The one you are constructing in LeetCode is Binary Tree having almost 2 children.

    Here you need to use Adjacency list, which can be implemented via vectors in C++. You should read about it.

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4 года назад, # |
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Any hint for problem F? (getting wrong answer)

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    4 года назад, # ^ |
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    Spoiler
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4 года назад, # |
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Thank you very much for this effort. I really liked the problems.

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4 года назад, # |
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I solved F by observing some examples, can somebody give proof for F, why this works?

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    4 года назад, # ^ |
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    if you supposed a minimal matching where there are only two nodes on each side , and you paired each node with the one in opposite side you'll find that you can minimize more the distances by pairing each node with the one on it's side.

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      4 года назад, # ^ |
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      "you can minimize more the distances by pairing each node with the one on its side."

      I am looking for why this is always true. Maybe here it is easy to observe.

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        4 года назад, # ^ |
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        Consider the graph: 1-2, 2-3, 2-4

        Node 1 and node 3 have one seed each. Let us consider that there is a pot in the subtree of node 4 (let's call this node A) and a pot in the rest part of the tree above of the node 1 (let's call this node B).

        Optimal solution is to match seed-3 with pot-A and seed-1 with pot-B, because the sum of distances is equal to dist(3, 2) + dist(2, A) + dist(1, B). If you try to "cross" the edge (1-2) and match the seeds in a different way, then the sum would be equal to dist(3, 2) + dist(2, 1) + dist(1, B) + dist(1, 2) + dist(2, A) which is larger than before by 2 * dist(1, 2).

        As you can see, you can never avoid some distances, so there is no reason to try to match a seed with an other pot when there is already one available in its subtree.

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    4 года назад, # ^ |
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    I'm failing to understand the input test cases in problem F, where queries are :

    1 6 2
    2 5 3
    4 3 0
    

    Above, 2 seeds at node 1 and 2 pots at node 6 make 2 valid pairs(path 2->6 goes through edge 3-4). Similarly, 3 seeds at node 2 and 3 pots at 5 make 3 valid pairs. This path goes through edge 3-4. This makes a total of 5 valid pairs.

    6 2 2
    4 3 0
    

    Similarly, 2 seeds at node 6 and 2 pots at node 2 make 2 valid pairs and this path connecting 2 and 6 also cross the edge 4-3. So, total of 5+2=7 valid pairs. But, why is the answer for that query 3?

    I'm missing something here, kindly help!!

    Also, it's written The total sum of the distances between every pair must be minimized to keep clean the air. Does this mean something? I didn't get it.

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      4 года назад, # ^ |
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      • It's not required that seeds should placed in the respective same pots which were created while creating seeds in each query.

      • In the above example for the last query we can use 2 pots at node 2 for placing seeds from node 1 or 2.

      The total sum of the distances between every pair must be minimized to keep clean the air

      This means that sum distance travelled by each seed from the source node to node which contains pot should be minimized.

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4 года назад, # |
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Will all the test cases be made public after the contest ends?

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4 года назад, # |
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I am getting TLE in TC 4. I stretched the tree along the diameter and after that if node is added to the end points of diameter then dynamic diameter will increase by 1 otherwise not. My solution is here. Please somebody tell me error in my approach.

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4 года назад, # |
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1 sec time limit is pretty rough for java?

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4 года назад, # |
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Can someone please explain me what's wrong in my logic for B ? I seem to be repeatedly failing test case #6.

Logic
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    4 года назад, # ^ |
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    This will not give you the correct answer.

    Spoiler
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4 года назад, # |
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Can anyone pls provide some test case for problem C? im getting WA on test 4

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4 года назад, # |
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In question no C these are the arrays in my programm my code is runnig is fine in my system btt shws runtime error when i submit can this be a memory issue??

int N, D;
int depth[300010];//depth of node
vector<int> adj[300010];
int par[20][300010];//binary lifting table

Inside main

    memset(depth, 0, sizeof(depth));
    for (int i = 0; i < 300010; i++)
    {
        adj[i].clear();
    }
    memset(par, -1, sizeof(par));
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4 года назад, # |
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SecondThread, what's the constraint of q in D?

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    4 года назад, # ^ |
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    Oh, good question. I’m pretty sure it’s 3*10^5. But like, answer each query in sublinear time please and you should be good to go.

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4 года назад, # |
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Thank you very much for the problemset, it's very helpful! I like how it's not just straightforward "implement LCA and be done with it", but includes some modification that forces you to understand the concept behind it.

I have a question about F. It seems that the observation is

Spoiler

Which seems intuitively true, but how to prove it?

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    4 года назад, # ^ |
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    Spoiler
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      4 года назад, # ^ |
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      Can i get a hint for problem F.

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        4 года назад, # ^ |
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        The problem is strictly linked with the idea of minimizing the sum of distances between seeds and pots, thus the first thing to think is "when is it necessary to cross edge $$$E$$$ when matching a seed with a pot?".

        This is not likely a hint, but let's try to see something intuitively in the following cases:

          R -> root.
         / \
        P   S
        

        Here you have a pot and a seed in the leaf nodes, thus the result in any edge would be 1 since there's no other matching possible/closer/minimum. This would give us a total sum of 2.

               __R__          -> root.
             /       \
            N1        N2      -> nodes.
          /  \       /  \
        2*P  1*S   1*P  2*S   -> amount of seeds/pots.
        N3   N4    N5   N6    -> nodes.
        

        Here our options would be:

        • Match all pots from $$$N3$$$ with all seeds in $$$N6$$$ and the seed in $$$N4$$$ with the pot in $$$N5$$$. This would give us a total distance sum of 4*2 + 4*1 = 12 and the following edge values:
             __R__         -> root.
           3/     \3       -> edge values.
           N1      N2      -> nodes.
         2/  \1  1/  \2    -> edge values.
        N3   N4  N5   N6   -> nodes.
        
        • Another option would be to match 1 pot from $$$N3$$$ with the seed in $$$N4$$$, the remaining pot from $$$N3$$$ with a seed in $$$N6$$$, the pot from $$$N5$$$ with the remaining seed in $$$N6$$$. This would give us an smaller total distance sum of 2*1 + 4*1 + 2*1 = 8 and the following edge values:
             __R__         -> root.
           1/     \1       -> edge values.
           N1      N2      -> nodes.
         2/  \1  1/  \2    -> edge values.
        N3   N4  N5   N6   -> nodes.
        

        Note that the second option is better because we don't need to cross the root edges multiple (extra) times. Thus, matching seeds/pots locally (closer matches) might reduce the total sum. Also, looking at the last example, try to answer the question "when is it necessary to cross edge $$$E$$$ when matching seeds and pots?" thinking locally.

        After solving (or if you still can't solve) the problem look at the spoilers on this thread to clear any doubt.

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          4 года назад, # ^ |
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          Thanks bro for your effort, I think i got the key now.

          My Idea
          UPD: Solved!, Thank you bro.
          
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4 года назад, # |
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I am not able to solve circumference of tree problem .. also there is no editorial available .. what to do ?

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4 года назад, # |
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It's really helpful! Thank you !

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4 года назад, # |
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Can some one plz provide a link to their submission for E. I have understood the approach (partially), but having problem to implement it.

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4 года назад, # |
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Great video, thank you!

Also, I think a thing worth mentioning is that segment trees can be directly utilized on incomplete rooted binary trees, while this tree -> Euler path -> segment tree approach helps us reduce the complexity when trees aren't necessarily binary.

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4 года назад, # |
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In problem E SecondThread, I got 4 WA's using Fenwick/BIT Tree but as soon as I changed it to Segment Tress, it got accepted.

Solution using BIT/Fenwick Tree

Solution using Segment Tree

I am unable to understand why it is happening. If anyone has solved this question using fenwick tree, please look at the code and tell where is it going wrong or share your code.

UPD: I got the error. I was constructing the fenwick tree in a wrong manner.

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    4 года назад, # ^ |
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    Hmm, my model solution uses a fenwick tree. That’s odd for sure...

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      4 года назад, # ^ |
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      If that's so, then I must be doing some silly/ gross mistake that I am unable to find. I hope you wouldn't mind to taking a peek at my code unless you're busy.

      Anyway, thanks for the clarification.

      UPD: I got the error, I was construting the Fenwick Tree in a wrong way.

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4 года назад, # |
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Nice guide, really appreciated.

For everyone looking for other problems to apply the tree flattening trick, here are some from CSES problem set:

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4 года назад, # |
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Can someone please let me know why My code fails on test 5(Problem B)

code

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4 года назад, # |
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For problem C: Assume tree like following.

7
1 2
1 3
2 6
2 7
3 4
3 5
2
7 5 1
7 5 3

Is the answer (2, 1) or (2, 3)?

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4 года назад, # |
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Can anyone tell me why I am getting the wrong answer for testcase-4 in problem E, or give me any test case? SecondThread

code

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4 года назад, # |
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Can i get hint for problem F.

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4 года назад, # |
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Can someone please explain me the way to approach problem F.I have been stuck at it for quite some time now !!

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4 года назад, # |
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can anyone tell how to do c

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4 года назад, # |
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Auto comment: topic has been updated by SecondThread (previous revision, new revision, compare).

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4 года назад, # |
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Really enjoyed thanks.

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4 года назад, # |
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SecondThreads java solutions were posted, but if you need A-E in c++: A B C D E F

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4 года назад, # |
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I'm facing a weird error. The same solution is getting accepted in C++14 and getting RE in C++17 for Problem B.

Accepted code

Same code is getting RE in C++17, I tried
1. using memset for dist arrays
2.return 0 for main fn

SecondThread What could be the problem?

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    4 года назад, # ^ |
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    That sounds more like a C++ issue to me than anything else. If I had to guess, I would say it's likely a stack-space configuration issue. I think the RTE is probably that your recursion goes too deep on the rope case for instance, which causes you to run out of stack space.

    I don't know what differences there would be otherwise, although I was under the impression that your main was always suppose to return 0...

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4 года назад, # |
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SecondThread the problems aren't available in the gym contest

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3 года назад, # |
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SecondThread if possible, maybe u can link the video editorial as editorial in the problemset. That may help a little bit

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4 недели назад, # |
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249461942 can someone please help for 102694B - Dynamic Diameter why is this failing for test case 6?