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nodchip
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Sep 27, 2026
| ## step1 | ||
| 30m以上考えて分からなかったので Hint として以下を見て解いた。合計で1h以上かかった。 | ||
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| k=2までは自力で考えられたが、k > 2 の答えに自力で辿り着けなかった。 |
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自分も自力でたどり着けませんでした。 k=2 の場合は、連鎖行列積問題に似ているように感じました。
https://en.wikipedia.org/wiki/Matrix_chain_multiplication
最後のマージの直前の状態を作るための最小をコストを先に求めたあと、マージのコストを追加する、という点がポイントのようですが、自力で発想できる気がしませんでした。
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連鎖行列積問題は初めて知りました。二つの行列をまとめる処理や動的計画法で解ける点は似ていそうです。
よく復習しようと思います。
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| prefix_sum = list(itertools.accumulate(stones, initial=0)) | ||
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| dp = [[[float("inf")] * (k + 1) for _ in range(num_stones)] for _ in range(num_stones)] |
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こちらのコメントをご参照ください。
Yuto729/leetcode#105 (comment)
min_cost あたりが良いと思います。
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min_cost としました。dp はなるべく使わないように気をつけます。
| dp[i][i][1] = 0 | ||
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| for length in range(2, num_stones + 1): | ||
| for i in range(num_stones - length + 1): |
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i と j の出現回数が多く、登場するたびに i と j が何を表すかを考える認知負荷が高そうだと感じました。 left と right あたりにしたほうが認知負荷が低くなりそうだと思いました。
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https://leetcode.com/problems/minimum-cost-to-merge-stones/description/