結果
問題 | No.484 収穫 |
ユーザー |
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提出日時 | 2025-03-20 18:48:37 |
言語 | PyPy3 (7.3.15) |
結果 |
WA
|
実行時間 | - |
コード長 | 1,489 bytes |
コンパイル時間 | 418 ms |
コンパイル使用メモリ | 82,356 KB |
実行使用メモリ | 540,760 KB |
最終ジャッジ日時 | 2025-03-20 18:50:24 |
合計ジャッジ時間 | 24,702 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 2 WA * 1 |
other | AC * 9 MLE * 12 |
ソースコード
n = int(input()) A = list(map(int, input().split())) INF = float('inf') dp = [[[INF] * 2 for _ in range(n)] for __ in range(n)] # Initialize for single-cell intervals for i in range(n): dp[i][i][0] = A[i] dp[i][i][1] = A[i] # Process intervals of increasing length for length in range(1, n): for l in range(n - length): r = l + length # Check if we can extend to the left (to l) # From the right end of [l+1][r] candidate_left_from_right = dp[l+1][r][1] + (r - l) new_time = max(candidate_left_from_right, A[l]) if new_time < dp[l][r][0]: dp[l][r][0] = new_time # From the left end of [l+1][r] candidate_left_from_left = dp[l+1][r][0] + 1 new_time = max(candidate_left_from_left, A[l]) if new_time < dp[l][r][0]: dp[l][r][0] = new_time # Check if we can extend to the right (to r) # From the left end of [l][r-1] candidate_right_from_left = dp[l][r-1][0] + (r - l) new_time = max(candidate_right_from_left, A[r]) if new_time < dp[l][r][1]: dp[l][r][1] = new_time # From the right end of [l][r-1] candidate_right_from_right = dp[l][r-1][1] + 1 new_time = max(candidate_right_from_right, A[r]) if new_time < dp[l][r][1]: dp[l][r][1] = new_time # The answer is the minimum of the two possible end states for the full interval [0, n-1] print(min(dp[0][n-1][0], dp[0][n-1][1]))