## Editorial for GlobeX Cup '18 J5 - Errands

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Submitting an official solution before solving the problem yourself is a bannable offence.

Author: Plasmatic

For the first subtask, it suffices to BFS from to and from to for each query. Then, to get the answer for the queries we can just sum the distance from to and to .

For this subtask, if is unreachable through either the first or second BFS, print This is a scam!.

Time Complexity:

For the second subtask, note that we can instead BFS from to and instead of the other way around. With that information in mind, we can also note that stays constant throughout the queries, so we only need to run one BFS at the beginning from to every other node.

Now, all we have to do for each query is sum and .

Note that if either or are unreachable, print This is a scam!.

Time Complexity: