Editorial for Yet Another XOR Tree Problem

Subtask 1:
Any bruteforce solution should pass.

Subtask 2:
In each query(U,V)query(U, V), treat node UU as the root node of the tree. Run a BFS/DFS from node UU and track the nodes that lie on the simple path between node UU and node VV. You can do it by managing a par[]par[] array, where par[i]=immediate parent of node ipar[i] = immediate\space parent\space of\space node\space i. You should also calculate F(U,i)F(U, i) for all node ii while running BFS/DFS. The rest is for the solver to solve.

Subtask 3:
If you’re not familiar with finding maximum KAiK \oplus A_iusing Trie data structure, read this at first.

Let’s take an arbitrary node as the root of the tree. In the below example, we’ll treat node 11 as the root node.

The weights of the edges are written in red colour. Here, F(1,4)=96, F(1,3)=95F(1, 4) = 9\oplus 6, \space F(1, 3) = 9\oplus 5. So, F(1,4)F(1,3)=9695=65=3F(1,4) \oplus F(1,3) = 9\oplus 6\oplus 9\oplus 5 = 6 \oplus 5 = 3, which is clearly the value of F(4,3)F(4, 3). That is to say, we can write F(U,V)F(U, V) as F(root,U)F(root,V)F(root, U) \oplus F(root, V).

Let, the Lowest Common Ancestor (LCA) of node UU and node VV is node LL. Now in queries, we can split the query(U,V)query(U, V) queries into two separate queries: query(L,U,F(root,U))query(L, U, F(root, U)) and query(L,V,F(root,U))query(L, V, F(root, U)), where query(x,u,k)query(x, u, k) means the maximum value of kF(root,i)k\oplus F(root, i) such that node ii lies on the simple path connecting node xx and node uuxx is node uu or any ancestor of node uu. The maximum of these two splitted queries will be our desired answer for the real query.

To answer query(x,u,k)query(x, u, k), we need to precalculate F(root,u)F(root, u) for all node uu at first. After that, we can build a trie only using the values of F(root,j)F(root, j), where jj is node uu or any ancestor of node uu. Then it comes down to the normal trie problem stated at the beginning of the editorial (subtask 3). To ensure that we don’t search in a trie node created by only the ancestors of node xx, we may track the maximum dfs discovery time in trie nodes or do similar kinds of things.

As building trie using the values of F(root,j)F(root, j) for each node uu will be a huge memory-consuming task, we need to use Persistent Trie data structure.

This problem can also be solved using a normal Trie data structure by answering the queries offline. The above concept is helpful in this approach too.

Setter’s Sol: https://pastebin.com/Xdn7SrDD

    Statistics


    75% Solution Ratio

    Um_nikEarliest, 3M ago

    user.794723Fastest, 0.7s

    user.794723Lightest, 88 MB

    Deshi_TouristShortest, 2724B

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