ID: fantasydraft

Difficulty: 4.1

CPU Time: 3 seconds

Memory: 1024 MB

This solution is not too difficult to understand the approach, but the implementation is a bit tricky.

Since there are n teams, and k players per team, there are n * k players to be assigned, which may be fewer than the total p players that are available. Thus, we loop k times, and each time loop n times, to assign the k^th^ player to the n^th^ team. At any team's turn, we can pop the first player off of the respective team's preference list, and if that player is available, assign it, and mark the player as no longer available. If that player is taken, repeat with the next preferred player, and try again, etc. If there is no preference list for this team (or it has been depleted), one can simply take the first-ranked player and assign it.

Since one might use a list / queue / vector to store available players, consider the extreme cases where this linear-search/extraction cost will take too long. Imagine the preference lists of all the teams being the lowest-ranked players. The preference lists of all teams being merged at each index, and the order reversed, consituting the tail of the preference list, and n * k being far lower than p; one has to traverse every single available player to find the respective team's player.

To achieve a faster lookup/extraction, one should consider something such as Maps to convert between a rank position and the player name to achieve much faster lookups.

Included is the solution made during the 2019 ICPC! Python sets saved us.