Assuming, we know the minimum cost required for every dish in meal i, we can transition to i+1. To transition, the nodes in i+1 will take the smallest costing dish in i if possible ( no bad edge between them ). We can therefore sort nodes in i based on their cost, then pair up the current node with all possible nodes in i+1. Repeat for all i(1,2,3).

I think this should be O(nlogn) because of sorting, though there is a large constant factor due to iterating on each node also being O(nlogn).