Блог пользователя desperateboy12344

I need help with problem D in the recent Deltix Round
Автор desperateboy12344, история, 3 месяца назад, По-английски

As my title stated, I really need help with understanding D (https://codeforces.me/contest/2002). (Sorry, I'm really dumb :<<<). So the statement is like this (D1): Given a permutation with some persistent swap queries and a dfs algorithm, check if the permutation after each swap can form a dfs order of some perfect binary tree. I really think that as long as the permutation starts with 1, we can always recreate a perfect binary tree from the permutation. Because like if we assign a pointer p = 0, when we traverse on a binary tree, when we enter a node u, we can just assign p = p + 1, and then assign a[u] = P[p] (P is the original permutation), hence it always exists (Did I misread the statement??). Help me guys I'm really dumb :<<<

  • Проголосовать: нравится
  • -2
  • Проголосовать: не нравится

»
3 месяца назад, # |
  Проголосовать: нравится 0 Проголосовать: не нравится

Don't apologize for being "dumb" — it's completely normal to have questions and misunderstandings! Let's break down the problem statement and your concerns:

Problem Statement: Given a permutation with some persistent swap queries, check if the permutation after each swap can form a DFS order of some perfect binary tree.

Your Concern: You think that as long as the permutation starts with 1, you can always recreate a perfect binary tree from the permutation. You're suggesting that by assigning a pointer p = 0 and traversing the binary tree, you can assign values to the nodes according to the permutation P.

Misunderstanding: The problem is not asking you to recreate a perfect binary tree from the permutation. Instead, it's asking you to verify whether the permutation after each swap can form a DFS order of some perfect binary tree. In other words, you need to check if the resulting permutation can be used as a DFS order of a perfect binary tree.

Key Point: The key point is that the DFS order of a perfect binary tree is not just about assigning values in the correct order. It's also about the structure of the tree. A perfect binary tree has a specific structure, where each node has at most two children (left and right). The DFS order should reflect this structure.

Correct Approach: To solve this problem, you need to check if the permutation after each swap can be used as a DFS order of a perfect binary tree. You can do this by iterating through the permutation and checking if each node has at most two children (left and right). You can also use a graph data structure to represent the binary tree and verify that it satisfies the properties of a perfect binary tree.

Hint: To simplify the problem, you can start by assuming that the permutation is initially a valid DFS order of a perfect binary tree. Then, you can apply the swap queries and check if the resulting permutation is still a valid DFS order of a perfect binary tree.

I hope this helps clarify things! If you have more questions or need further guidance, feel free to ask.

→ Ответить
  • »
    »
    3 месяца назад, # ^ |
      Проголосовать: нравится 0 Проголосовать: не нравится

    First of all thank you for answering my question.

    But I still don't really grasp it. In this example (https://codeforces.me/contest/2002/problem/D1)

    they say 1 3 5 4 2 6 7 can't be a dfs order of some perfect binary tree. But what about this tree

    Capture

    Anyway thanks for your time!!

    → Ответить
    • »
      »
      »
      3 месяца назад, # ^ |
        Проголосовать: нравится 0 Проголосовать: не нравится

      They're specifically talking about a perfect binary tree in which the parent of $$$i$$$ is $$$\left\lfloor\frac{i}{2}\right\rfloor$$$ for each $$$i$$$. This isn't true for your tree.

      → Ответить
  • »
    »
    3 месяца назад, # ^ |
      Проголосовать: нравится +7 Проголосовать: не нравится

    if you ever feel useless, remember that there are someone copy pasting people's questions into chatgpt and pasting the answer without changing a single word

    → Ответить