Hello,
This is working. You have two flights from B to C... And you can do A -> B -> C with the first and the last flight. x)

It could not work if you take max_layover_time > 12,
But as said Cédric, we won't have any scenarios like that (you can see it in "Some clarifications" post) and you can keep this condition because it's not that so long to check it and it can even be accelerated. Don't worry about pathological cases because this won't happen in scenarios we have ! ;)
It's hope to you to remove this function but :
- You won't have the same input as we have,
- You algorithm without a pathological case like that will be slower because it will create stupid loops.

I hope it answers to you,

Regards,

Don't worry, the algorithm is maybe not perfect for this very complicated problem ! =)
But Cédric will repeat what I said :
The scenarios that they give us will not have this kind of pathological cases, the goal of the contest is to make us accelerating the algorithm ;
Not correct it but only make it as fast as possible !

Regards,

As we already said on an other topic, you are free to implement your algorithm the way you want, but you must stick to the output that our reference code would give.

You have to admit that in real life, you would not flight through the same city twice just to get a discount.

Anyway, this is important for your result to check that from A to B (if A and B are your points of interest), you never fly twice to the same city.

EDIT :

I didn't really watched your solution, but the best flight seems to be :

A -> B, begin: hour 6, land: hour 10, cost $1

B -> C, begin: hour 10, land: hour 11, cost $1

With a cost = 2$. And as the min_time is 0 and max_time is 12, the default program should find this solution.

You don't need to wait for 6 hours in A, you just have to start your trip on the flight A -> B, begin: hour 6, land: hour 10, cost $1 since your travel is from A to B.

But anyway, you seem to have your answer.