3 ms·
Not as complicated as you might think. You essentially add time as an extra dimension to your graph, true, but because you're dealing with discrete vehicles ins
by BorgHunter 12y ago
Not as complicated as you might think. You essentially add time as an extra dimension to your graph, true, but because you're dealing with discrete vehicles instead of a continuously varying function, you can take shortcuts that make that portion of it pretty simple. It's the walk to and from the transit stops, in my experience, that's a PITA to deal with when you're trying to do transit routing.
- michaelt 12y agoIf you're doing simple Dijkstra and you only want to return a result for a single departure time, you're right that it's not too complicated as you still only have one cost per edge. If you want to do computations in advance, as you need to for fast algorithms like Contraction Hierarchies, you have a cost-vs-time profile for each edge. Things can get complicated quickly.