Editorial for DMOPC '18 Contest 6 P3 - Wish Upon a Star - Olympiads Online Judge

Editorial for DMOPC '18 Contest 6 P3 - Wish Upon a Star

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: Kirito

For $20\%$ ~20\%~ of points, we can add each edge, and see if it forms a cycle (namely, it connects two already connected nodes). We count the number of such edges, and output YES if there is at most 1 such edge, and NO otherwise.

Time Complexity: $\mathcal O(M^2)$ ~\mathcal O(M^2)~

Let $C$ ~C~ denote the number of connected components. This can be computed in $\mathcal O(N + M)$ ~\mathcal O(N + M)~ using DFS.

For $100\%$ ~100\%~ of points, we observe that the answer is always yes if $M = N - C$ ~M = N - C~ or $M = N - C + 1$ ~M = N - C + 1~. The first case gives us a forest, and the second case forms a forest with a single extra edge.

Time Complexity: $\mathcal O(N + M)$ ~\mathcal O(N + M)~

Comments

1

sre0x2a commented on June 23, 2025, 7:08 p.m.

I am a little puzzled why a Disjoint Set would not be the recommended approach to solving this problem.

2

Kirito commented on June 24, 2025, 11:26 p.m.

DFS is conceptually simpler and more likely to be known by those reading the editorial.