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nodchip
reviewed
Nov 10, 2025
| - 空間計算量はO(min(m, n))になる | ||
| - スタート地点に障害物があるケースを考えると、DP[0][0]の初期化も障害物があるか確認したほうがよい | ||
| - 前回の問題ではnCrで解けたが、今回は無理そう | ||
| - 他の解き方が出てこないので、読みやすい二次元DPで解く。 |
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上から下に見ていって、障害物がない行については、ある行の結果のベクトルに、以下のような行列を掛けてあげることで、結果を求めることができます。
1 1 1 1 1 ...
0 1 1 1 1 ...
0 0 1 1 1 ...
0 0 0 1 1 ...
0 0 0 0 1 ...
...
DP の計算式を行列の掛け算で表しています。
また、障害物がない行が M 行続く場合、行列の M 乗を計算してからベクトルに掛けることで、高速化できます。 N × N の行列の掛け算は O(N^3) で求められるので、 M 乗は Exponentiation by Squaring で O(N^3 log M) で求めることができます。
障害物がある行は DP で求められます。
ソフトウェアエンジニアの常識には入っていないと思います。 Exponentiation by Squaring の部分は、単体では常識に含まれていると思います。
naoto-iwase
reviewed
Nov 11, 2025
Comment on lines
+88
to
+90
| if (column > 0) { | ||
| num_paths[column] += num_paths[column - 1]; | ||
| } |
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自分はif column == 0: continueで弾いてからnum_paths[column] += num_paths[column - 1]と書きたくなるんですが、好みの範囲だと思います。
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今回の問題
https://leetcode.com/problems/unique-paths-ii/
次の問題
https://leetcode.com/problems/house-robber/