One-way flight trip problem

Question

You are going on a one-way indirect flight trip that includes billions an unknown very large number of transfers.

• You are not stoppi

Answer 1

Each airport is a node. Each ticket is an edge. Make an adjacency matrix to represent the graph. This can be done as a bit field to compress the edges. Your starting point will be the node that has no path into it (it's column will be empty). Once you know this you just follow the paths that exist.

Alternately you could build a structure indexable by airport. For each ticket you look up it's src and dst. If either is not found then you need to add new airports to your list. When each is found you set a the departure airport's exit pointer to point to the destination, and the destination's arrival pointer to point to the departure airport. When you are out of tickets you must traverse the entire list to determine who does not have a path in.

Another way would be to have a variable length list of mini-trips that you connect together as you encounter each ticket. Each time you add a ticket you see if the ends of any existing mini-trip match either the src or dest of you ticket. If not, then your current ticket becomes it's own mini-trip and is added to the list. If so then the new ticket is tacked on to the end(s) of the existing trip(s) that it matches, possibly splicing two existing mini-trips together, in which case it would shorten the list of mini-trips by one.