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My first instinct was just run Dijkstra and call it a day, which is obviously wrong.
Model the problem as a shortest path in an augmented state space where the state is (node, number of extra edges used). Use Dijkstra's algorithm on this expanded graph, where original edges have their given weights and extra edges of weight 1 can be added between any pair of nodes. The answer is the minimum distance to (last node, k) for any k ≤ K.
Pro tip: Emphasize that adding an extra edge between any two nodes is equivalent to allowing a transition from any node to any other node with cost 1, which can be handled efficiently by maintaining a global minimum distance for each usage count. This avoids explicitly adding O(V^2) edges and keeps the solution scalable.
Create states (v, k) where v is a node and k is the number of extra edges used so far (0 ≤ k ≤ K). This captures the trade-off between using original edges and extra edges.
From state (u, k), you can traverse an original edge (u, v) with weight w to reach (v, k), or use an extra edge to reach (v, k+1) with weight 1 for any v ≠ u, provided k < K.
Run Dijkstra on the expanded graph to find the shortest path from (1, 0) to any state (N, k) with k ≤ K. Use a priority queue and process states in increasing distance order.
Instead of iterating over all possible target nodes for extra edges, maintain for each k the minimum distance among all nodes at that k. Then an extra edge from any node at level k can reach any node at level k+1 with cost 1 plus that minimum, reducing complexity.
After Dijkstra, the answer is the minimum distance among all states (N, k) for k = 0 to K. If no path exists, return -1 or infinity as appropriate.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.