結果
| 問題 |
No.484 収穫
|
| コンテスト | |
| ユーザー |
lam6er
|
| 提出日時 | 2025-04-16 00:11:21 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 1,298 bytes |
| コンパイル時間 | 302 ms |
| コンパイル使用メモリ | 81,672 KB |
| 実行使用メモリ | 526,064 KB |
| 最終ジャッジ日時 | 2025-04-16 00:13:07 |
| 合計ジャッジ時間 | 23,360 ms |
|
ジャッジサーバーID (参考情報) |
judge5 / judge1 |
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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')
# Using 1-based indices for cells (1 to n)
# dp[l][r][0] is the time when covering [l..r] ending at l
# dp[l][r][1] is the time when covering [l..r] ending at r
dp = [[[INF] * 2 for _ in range(n + 2)] for __ in range(n + 2)]
for i in range(1, n + 1):
dp[i][i][0] = a[i - 1]
dp[i][i][1] = a[i - 1]
for length in range(2, n + 1):
for l in range(1, n - length + 2):
r = l + length - 1
# Calculate dp[l][r][0] (ending at left end l)
# Option 1: come from [l+1..r] ending at left (l+1)
option1 = dp[l + 1][r][0] + 1
# Option 2: come from [l+1..r] ending at right (r)
option2 = dp[l + 1][r][1] + (r - l)
dp[l][r][0] = min(option1, option2)
dp[l][r][0] = max(dp[l][r][0], a[l - 1]) # Ensure we can harvest at l
# Calculate dp[l][r][1] (ending at right end r)
# Option 1: come from [l..r-1] ending at right (r-1)
option1 = dp[l][r - 1][1] + 1
# Option 2: come from [l..r-1] ending at left (l)
option2 = dp[l][r - 1][0] + (r - l)
dp[l][r][1] = min(option1, option2)
dp[l][r][1] = max(dp[l][r][1], a[r - 1]) # Ensure we can harvest at r
answer = min(dp[1][n][0], dp[1][n][1])
print(answer)
lam6er