結果
| 問題 |
No.1369 交換門松列・竹
|
| コンテスト | |
| ユーザー |
lam6er
|
| 提出日時 | 2025-04-16 15:41:13 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 3,167 bytes |
| コンパイル時間 | 252 ms |
| コンパイル使用メモリ | 82,048 KB |
| 実行使用メモリ | 79,360 KB |
| 最終ジャッジ日時 | 2025-04-16 15:45:04 |
| 合計ジャッジ時間 | 3,402 ms |
|
ジャッジサーバーID (参考情報) |
judge4 / judge1 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 1 |
| other | AC * 23 WA * 10 |
ソースコード
import sys
def is_kadomatsu(a, b, c):
if a == b or a == c or b == c:
return False
return (b == max(a, b, c)) or (b == min(a, b, c))
def get_affected_triplets(i, n):
triplets = []
# Check triplet starting at i-2
s = i - 2
if s >= 0 and s + 2 < n:
triplets.append(s)
# Check triplet starting at i-1
s = i - 1
if s >= 0 and s + 2 < n:
triplets.append(s)
# Check triplet starting at i
s = i
if s + 2 < n:
triplets.append(s)
return triplets
def solve():
input = sys.stdin.read().split()
ptr = 0
T = int(input[ptr])
ptr += 1
for _ in range(T):
N = int(input[ptr])
ptr += 1
A = list(map(int, input[ptr:ptr+N]))
ptr += N
# Find all invalid triplets
S = []
for s in range(N - 2):
a, b, c = A[s], A[s+1], A[s+2]
if not is_kadomatsu(a, b, c):
S.append(s)
# If there are more than 4 invalid triplets, it's impossible
if len(S) > 4:
print("No")
continue
# Collect all positions in invalid triplets
positions = set()
for s in S:
positions.add(s)
positions.add(s + 1)
positions.add(s + 2)
positions = list(positions)
found = False
# Check all pairs of positions
for i in range(len(positions)):
for j in range(i + 1, len(positions)):
x = positions[i]
y = positions[j]
if A[x] == A[y]:
continue
# Check if all triplets in S contain x or y
valid = True
for s in S:
triplet_pos = [s, s+1, s+2]
if x not in triplet_pos and y not in triplet_pos:
valid = False
break
if not valid:
continue
# Perform swap
A[x], A[y] = A[y], A[x]
# Collect all triplets to check
triplets_to_check = set()
# Check triplets involving x
for s in get_affected_triplets(x, N):
triplets_to_check.add(s)
# Check triplets involving y
for s in get_affected_triplets(y, N):
triplets_to_check.add(s)
# Check all these triplets
ok = True
for s in triplets_to_check:
a = A[s]
b_val = A[s+1]
c = A[s+2]
if not is_kadomatsu(a, b_val, c):
ok = False
break
if ok:
found = True
# Undo swap before returning
A[x], A[y] = A[y], A[x]
print("Yes")
break
# Undo swap
A[x], A[y] = A[y], A[x]
if found:
break
if found:
continue
else:
print("No")
if __name__ == "__main__":
solve()
lam6er