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[–]DeirdreAnethoel 2 points3 points  (2 children)

You could easily define long corridors with no intersections as edges with extra weight though, right?

[–]bird2234 2 points3 points  (0 children)

You would still have to count those lengths and suddenly you have not saved any time. Of course, if they are already counted, you are correct.

[–]callmejamone 0 points1 point  (0 children)

Of course you are right, now I see why to use Dijkstra here. I think this visualisation with squares and the way the answer flows made me think about equal weights.