結果
| 問題 |
No.2036 Max Middle
|
| コンテスト | |
| ユーザー |
lam6er
|
| 提出日時 | 2025-03-31 17:47:12 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
TLE
|
| 実行時間 | - |
| コード長 | 1,699 bytes |
| コンパイル時間 | 228 ms |
| コンパイル使用メモリ | 83,020 KB |
| 実行使用メモリ | 297,832 KB |
| 最終ジャッジ日時 | 2025-03-31 17:48:02 |
| 合計ジャッジ時間 | 4,369 ms |
|
ジャッジサーバーID (参考情報) |
judge5 / judge3 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 4 |
| other | AC * 4 TLE * 1 -- * 12 |
ソースコード
import sys
from collections import deque
def main():
N_and_rest = list(map(int, sys.stdin.read().split()))
N = N_and_rest[0]
A = N_and_rest[1:N+1]
initial_peaks = []
for i in range(N - 2):
if A[i] < A[i + 1] and A[i + 1] > A[i + 2]:
initial_peaks.append(i)
in_queue = [False] * (N - 2)
q = deque()
for i in initial_peaks:
q.append(i)
in_queue[i] = True
count = 0
while q:
i = q.popleft()
in_queue[i] = False
# Check if the current triplet is still a peak
if i + 2 >= N:
continue
a = A[i]
b = A[i + 1]
c = A[i + 2]
if a < b and b > c:
# Process the peak
min_val = min(a, c)
new_b = min_val - 1
A[i + 1] = new_b
count += 1
# Check left neighbor triplet (i-1)
left_i = i - 1
if left_i >= 0 and left_i < (N - 2):
if not in_queue[left_i]:
is_peak = (A[left_i] < A[left_i + 1] and A[left_i + 1] > A[left_i + 2])
if is_peak:
q.append(left_i)
in_queue[left_i] = True
# Check right neighbor triplet (i+1)
right_i = i + 1
if right_i < (N - 2):
if not in_queue[right_i]:
is_peak = (A[right_i] < A[right_i + 1] and A[right_i + 1] > A[right_i + 2])
if is_peak:
q.append(right_i)
in_queue[right_i] = True
print(count)
if __name__ == "__main__":
main()
lam6er