Shortest Path In A Grid With Obstacles Elimination - #157
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Aug 10, 2026
| step = (ii, jj, k-grid[ii][jj], s+1) | ||
| if step[0:3] not in seen: | ||
| seen.add(step[0:3]) | ||
| dq.append(step) |
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これ、ここまでの破壊回数が1で到達できたならば、破壊回数が2では調べる必要はないのではないでしょうか。
私はこれはダイクストラに破壊回数がついていて、到達時間と破壊回数でパレート最適なのだけ残すというイメージです。
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dequeを使ったBFS実装も書いてみました。辺コスト1なのでダイクストラと同等です。破壊回数最大だけを残すのはパレート最適といえますね。
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https://leetcode.com/problems/shortest-path-in-a-grid-with-obstacles-elimination/description/?envType=problem-list-v2&envId=7p55wqm