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CAPIMOVE - Editorial

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Author:Maksym Bevza
Tester:Arjun Arul
Editorialist:Misha Chorniy

Difficulty:

Simple

Pre-Requisites:

none

Problem Statement

You are given tree consisting $N$ nodes. Each node $i$ has value $P_{i}$ assigned to it. All values $P_{i}$ are distinct. For each node $j$ find maximal value of node in the graph if to remove nodes adjacent to $j$ and $j$.

Explanation

Subtask 1 and 2

$N$ is less or equal than 10000. For each node will mark all nodes which are adjacent to it as dead(not usable), will find maximum value between non-marked nodes.

for i= 1..N //Step 1 for j=1..N alive[j] = 1; //initialize all marks as ones //Step 2 alive[i] = 0; //mark itself as dead for v is adjacent to i //mark all adjacent nodes alive[v] = 0; //mark neighbours as dead ans[i] = 0; //Step 3 for j=1..N if alive[j]==1 //if node j is alive ans[i]= max(ans[i],p[j]); //find maximum value

How long it works? Obviously, the complexity of first and third step is $O(N)$, how to find the complexity of the second step? Denote $D(v)$ - degree of node $v$, for vertex $i$ complexity of step 2 is equal to $D(i)$, over all iterations complexity of second step is $D(1)$+$D(2)$+..+$D(N)$, what is $2*(N-1)$ for trees, we see that each edge using exactly two times. Total complexity of this algorithm is $O(N*N+2*(N-2)+N*N)$ = $O(N^2)$

First and third steps are slowest in the code above, how to optimize it? We can rewrite that code a bit. for i=1..N alive[i]=1; for i=1..N alive[i]=0; for v is adjacent to i alive[v] = 0; ans[i]=0; for j=1..N if alive[j] == 1 //if node j is non-marked ans[i] = max(ans[i],p[j]); //update answer alive[i]=1; for v is adjacent to i alive[v] = 1;

Now total complexity is $O(N+2*(N-2)+N*N+2*(N-2))$ = $O(N^2)$, still $O(N^2)$, we can make some observations, basically we have set, where the following operations can be performed:

  • $alive_{i}=1$ is equivalent to adding in the set value of $p_{i}$
  • $alive_{i}=0$ is equivalent to erasing in the set value of $p_{i}$
  • Subtask 3

    Let's use some data structure, namely in our case, the data structure which can add/erase elements, and find the maximal value in the set of these elements, but does this thing in time less than $O(N)$.

    for i=1..N add(P[i]); for i=1..N erase(P[i]); for v is adjacent to i erase(P[v]); ans[i] = getMaximalValue(); add(P[i]); for v is adjacent to i add(P[v]); What is the best data structure for these things, in most of the modern programming languages exists built-in data structures for such things, in C++ you can use set, Java has Set, Python has similar data structures. If your language doesn't have built-in data structures, you can read about heaps (http://pages.cs.wisc.edu/~vernon/cs367/notes/11.PRIORITY-Q.html) and realize it.

    Let's write code with the map in C++, the similar code can be written in Java with Map or in Python with the dictionary. set < int > S; for i=1..N // Step 0 S.insert(P[i]); //Initialize set //Adding element P[i] in the set for i=1..N //Step 1 S.erase(P[i]); //Erase value of node i from set for v is adjacent to i //Iterate over neighbours of node i S.erase(P[v]) //Erase values of neighbours of i //Step 2 if !S.empty() ans[i] = *S.rbegin(); //Value of greatest element //Step 3, rollback of the step 1 S.insert(P[i]); //Insert value of node i from set for v is adjacent to i //Iterate over neighbours of node i S.insert(P[v]) //Insert values of neighbours of i

    Complexity of adding/erasing of number in such data structures is $O(log N)$, summary complexity of the first steps is $O(N log N)$, the same is for third steps, Total complexity is $O(N log N + N log N + N log N + N log N)$ = $O(N log N)$

    The overall time complexity of this approach is $O(N log N)$.

    Solution:

    Setter's solution can be found here
    Tester's solution can be found here

    Please feel free to post comments if anything is not clear to you.


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