E - Stronger Takahashi Editorial by en_translator
This problem can be solved with a 01-BFS (Breadth-First Search).
Assuming that he does not punch until needed, we can assume that a \(2 \times 2\) region of destroying blocks is adjacent to the square Takahashi is at.
When Takahashi is in the square labeled \(T\) in the diagram below, after a single punch he can move to any square labeled *
, regardless of the previous state of the squares.
.***. ***** **T** ***** .***.
Therefore, we can assume that a move on foot to an adjacent square costs \(0\) and moving to any *
square after a punch costs \(1\), and perform a 01-BFS to solve the problem. The computational complexity is \(O(HW)\).
Similar problem: ABC176D “Wizard in Maze”
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