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Seems to be similar to a scheduling problem, which I have no experience with. How do you compare the weights between each mercenary while keeping in mind that you cannot alternate back to a previous mercenary?
I have a hunch that this would be vaguely similar to running a modified Dijkstra's algorthm on an singlely linked list that is absurdly interconnected with K starting nodes and K ending nodes.
So far, I've only made a basic solution that always chooses the lowest option available, but that does not take into account that a mercenary may have a better streak of low cost options latter on.