Query Based Rerooting Technique

Правка en40, от DeadlyCritic, 2020-04-19 00:02:14

Halo :D!

Story: It was 4AM and i was unable to sleep so i brought you a nice trick and couple of problems to practice it.

Introduction

(Our dp changes if we change root of the tree, otherwise it wont make any sens to use this trick)

Lets say we want to find $$$dp_v$$$ for every vertex in a tree, this dp must be updated using the children of vertex $$$v$$$. Then the trick allows you to move the root from a vertex to one of it's adjacent vertices in $$$O(1)$$$.

It will be much better if the problem wants you to move the root to all vertices and find maximum of dp-s or etc.

Implementation

First calculate dp for a root $$$rat$$$. Its obvious that if we change root $$$r$$$ to one of its adjacent vertex $$$v$$$, then only $$$dp_r$$$ and $$$dp_v$$$ can change, so that we can update $$$dp_r$$$ using its new children and past $$$dp_r$$$, also $$$dp_v$$$ can be updated the same way.

Note: Don't iterate over $$$r$$$ and $$$v$$$ children, it will be $$$O(\sum\limits_v\binom{d_v}{2})$$$ and $$$W(n^2)$$$ so just don't.($$$d_v$$$ is the degree of vertex $$$v$$$)

Now being able to move the root with distance one, we will run a dfs from $$$rat$$$, and move the root with dfs.

See the semi-code below, i hope it helps for better understanding:

Semi-code

Problems

I cant open some sites so there is no solution for them, sorry.

You can read the same technique here, also the site has some problems.

Problem1 Solution

Problem2

Problem3 Solution

Problem4

Problem5//no solution right now

Probelm6//no solution right now

Please comment the problems which can be solved using this trick.

Advanced

Lets say that the problem has queries, such that each query gives two vertices $$$r$$$ and $$$v$$$ and wants you to calculate $$$f(r, v)$$$, its equal to $$$dp_v$$$ when the root is $$$r$$$. It can be solved with rerooting + ancestors binary lifting, it will answers each query in $$$O(Q*log_2n)$$$ time($$$O(log_2n)$$$ for binary lifting), and also $$$O(n*log_2n)$$$ precalculation.

So lets see how it works, for every edge($$$v \to u$$$) find $$$dp_v$$$ if the root is $$$u$$$ and $$$dp_u$$$ if the root is $$$v$$$, it will take the time it takes to move the root around and an additional $$$O(n)$$$ memory. Then if the query is in form $$$v$$$ and $$$v$$$(i.e. it wants to find $$$dp_v$$$ such that the tree is rooted at vertex $$$v$$$ itself) then the answer will be $$$n$$$, otherwise the problem is reduced to the following problem:

You are given two vertices $$$r$$$ and $$$u$$$$$$(u \ne r)$$$ what is the second vertex in the unique path from $$$u$$$ to $$$r$$$(it means we want to find out which child of $$$u$$$ is the closest to $$$r$$$).

This problem can be solved using binary lifting, if $$$u$$$ is not an ancestor of $$$r$$$, then the answer is $$$u$$$'s parent(the root is vertex 1), otherwise move $$$r$$$ using binary lifting until it's depth is equal to $$$u$$$'s depth plus 1, it can be done in $$$O(log_2n)$$$, let the answer to the reduced problem be $$$z$$$, then the answer to the whole query is equal to $$$f(z, u)$$$ such that $$$z$$$ is adjacent to $$$v$$$, we have precalculated this value so just print it :).

See the semi-code below, i skipped known parts so i will use $$$moveto(u)$$$ to move the root to vertex $$$u$$$, $$$f(r, u)$$$ for $$$dp_u$$$ when the root is $$$r$$$ and $$$rise(r, u)$$$ to find the ancestor of $$$r$$$ with depth equal to $$$depth_u+1$$$ in $$$O(log_2n)$$$.

Semi-code

I tried to add a problem from polygon for the advanced part, the problem is completed but i couldn't bring it to codeforces, so you cant submit your code, but i give you the link for problem's package, it includes 50 tests, validator and checker, main correct solution and problem statement, you can try running them in polygon.Windows Package(50 MB) Linux Package(45 MB)

In the problem mentioned above $$$dp_u$$$ is the size of the subtree of $$$u$$$, so solution will be like this :

Solution

I used maps to store dp, so the solution is a little slower than expected. The solution is checked using stresses tab in polygon.

I hope it will help you with solving DP-Tree problems.

Thanks for your attention.

Теги #dp, tree_dp, trick, #dfs and similar, rerooting technique, persistent structures, persistent, #data structure

История

 
 
 
 
Правки
 
 
  Rev. Язык Кто Когда Δ Комментарий
en48 Английский DeadlyCritic 2022-01-09 07:09:59 258 Tiny change: 'n\n[link](codeforces' -> 'n\n[link](https://codeforces'
en47 Английский DeadlyCritic 2020-07-06 21:41:48 231
en46 Английский DeadlyCritic 2020-05-24 09:45:09 756
en45 Английский DeadlyCritic 2020-04-19 17:30:36 139
en44 Английский DeadlyCritic 2020-04-19 00:20:52 72
en43 Английский DeadlyCritic 2020-04-19 00:12:43 80
en42 Английский DeadlyCritic 2020-04-19 00:10:39 218
en41 Английский DeadlyCritic 2020-04-19 00:04:22 56
en40 Английский DeadlyCritic 2020-04-19 00:02:14 38
en39 Английский DeadlyCritic 2020-04-19 00:00:38 10 Tiny change: 'this will calculate move the ' -> 'this will move the '
en38 Английский DeadlyCritic 2020-04-18 23:59:56 15
en37 Английский DeadlyCritic 2020-04-18 23:58:30 9
en36 Английский DeadlyCritic 2020-04-18 23:56:03 2 Tiny change: 'ry edge($v\tou$) find $' -> 'ry edge($v \to u$) find $'
en35 Английский DeadlyCritic 2020-04-18 18:32:50 66
en34 Английский DeadlyCritic 2020-04-18 18:30:04 4 Tiny change: 'ry edge($v$ \to $u$) find $' -> 'ry edge($v\tou$) find $'
en33 Английский DeadlyCritic 2020-04-18 18:29:17 1 Tiny change: '$O(log_2n) for binar' -> '$O(log_2n)$ for binar'
en32 Английский DeadlyCritic 2020-04-18 18:19:01 86
en31 Английский DeadlyCritic 2020-04-18 18:16:23 8999
en30 Английский DeadlyCritic 2020-04-18 18:00:57 99
en29 Английский DeadlyCritic 2020-04-18 17:50:12 484
en28 Английский DeadlyCritic 2020-04-18 17:42:25 688
en27 Английский DeadlyCritic 2020-04-18 17:09:46 97
en26 Английский DeadlyCritic 2020-04-18 12:25:14 136
en25 Английский DeadlyCritic 2020-04-17 23:52:30 238
en24 Английский DeadlyCritic 2020-04-17 23:37:39 77
en23 Английский DeadlyCritic 2020-04-17 23:35:38 34 Tiny change: 'e problems, the implementation will be added.\n\nThank' -> 'e problems.\n\nThank'
en22 Английский DeadlyCritic 2020-04-17 23:35:11 1642
en21 Английский DeadlyCritic 2020-04-17 19:02:21 4
en20 Английский DeadlyCritic 2020-04-17 15:27:10 13
en19 Английский DeadlyCritic 2020-04-17 14:56:11 307 Tiny change: 'nd $u$$(u != v)$ what ' -> 'nd $u$$(u \ne v)$ what '
en18 Английский DeadlyCritic 2020-04-17 14:47:28 1599
en17 Английский DeadlyCritic 2020-04-17 12:42:25 3 Tiny change: '/abc160_f)\n///i cant o' -> '/abc160_f)//i cant o'
en16 Английский DeadlyCritic 2020-04-17 12:41:50 110
en15 Английский DeadlyCritic 2020-04-17 12:39:01 18 Tiny change: 'e way.\n\n$\binom{4}{2}$\n\n*Note:' -> 'e way.\n\n*Note:'
en14 Английский DeadlyCritic 2020-04-17 12:38:07 252 Tiny change: 'e way.\n\n*Note:' -> 'e way.\n\n$\binom{4}{2}$\n\n*Note:'
en13 Английский DeadlyCritic 2020-04-17 12:27:58 284
en12 Английский DeadlyCritic 2020-04-17 12:16:32 290
en11 Английский DeadlyCritic 2020-04-17 05:05:12 147
en10 Английский DeadlyCritic 2020-04-17 05:01:51 94
en9 Английский DeadlyCritic 2020-04-17 04:46:45 66 Tiny change: 'tions\n\n[halo](https://' -> 'tions\n\n[Tree Painting](https://'
en8 Английский DeadlyCritic 2020-04-17 03:54:24 601
en7 Английский DeadlyCritic 2020-04-17 03:46:53 20 Tiny change: ' Hi,\n Its 4AM ' -> ' Hi,\n Its 4AM '
en6 Английский DeadlyCritic 2020-04-17 03:43:46 20
en5 Английский DeadlyCritic 2020-04-17 03:41:57 12 Tiny change: 'low me to right it in wor' -> 'low me to --right-- write it in wor'
en4 Английский DeadlyCritic 2020-04-17 03:11:25 35
en3 Английский DeadlyCritic 2020-04-17 03:08:14 2 Tiny change: 'rks in $O(t)$ to calc' -> 'rks in $O(T)$ to calc'
en2 Английский DeadlyCritic 2020-04-17 03:07:48 156 (published)
en1 Английский DeadlyCritic 2020-04-17 03:05:35 2061 Initial revision (saved to drafts)