Editorial for ICPC NWERC 2011 C - Movie Collection

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.

Note, since $m, r = 100\,000$ $m, r = 100 000$ , you cannot update the stack in $\mathcal O(m)$ $O (m)$ time
Therefore you need some smart data structure to store information
Binary indexed tree/Fenwick tree does the job in $\mathcal O(\log m)$ $O (\log m)$ time
At every step, take movie $x$ $x$ and put it back on $-1, -2, -3, \dots$ $- 1, - 2, - 3, \dots$

Note that this is also $\mathcal O(n^2)$ $O (n^{2})$ :

Copy

for (int j = 0; j < r; j++) {
  movie = sc.nextInt();
  index = movies.indexOf(movie);
  System.out.print("" + (m - index - 1) + " ");
  movies.removeElementAt(index);
  movies.add(movie);
}

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