Editorial for CCC '15 S3 - Gates

Remember to use this editorial only when stuck, and not to copy-paste code from it. Please be respectful to the problem author and editorialist.

Submitting an official solution before solving the problem yourself is a bannable offence.

Author: FatalEagle

A simple greedy algorithm goes like this:

For each plane, assign it to the gate with the largest number that isn't occupied.

It's not hard to see why this is correct — indeed, intuitively, we're using the "worst" gate we can for each plane (since a gate with a lower number can serve at least every plane a gate with a higher number can).

Implementing this naïvely will result in a complexity of \mathcal{O}(GP), which is enough for the weak data used on the actual CCC. To speed this up, we can use a balanced binary search tree to store free gates. In C++, this is the set data structure, with a final time complexity of \mathcal{O}(P \log G).

Solution — C++

#include <stdio.h>
#include <bitset>

int a,b,c,d;
std::bitset<100005> taken;

int main()
    for (int i=0; i<b; i++) {
        if (d>a) {
            return 0;


  • 7
    tempusertest  commented on Dec. 11, 2020, 12:52 a.m.

    to confirm, even though the sample solution passes, it is demonstrating a naive (O(GP)) solution right? can't seem to find official documentation on this "._Find_next()" function, but from what ive seen on miscellaneous websites, it runs in O(n) (or something like O(n/32)) right?

    was confused because i inferred that an O(Plog(G)) solution was needed to pass on dmoj

  • -7
    bigboy2  commented on Dec. 10, 2020, 5:48 p.m. edited

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    • 8
      a_sad_javamain  commented on Dec. 10, 2020, 6:48 p.m. edited

      O(P(2logG)) = O(PlogG) since we disregard the constant factor