結果
| 問題 |
No.859 路線A、路線B、路線C
|
| コンテスト | |
| ユーザー |
lam6er
|
| 提出日時 | 2025-03-20 20:56:21 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 1,478 bytes |
| コンパイル時間 | 152 ms |
| コンパイル使用メモリ | 82,304 KB |
| 実行使用メモリ | 53,752 KB |
| 最終ジャッジ日時 | 2025-03-20 20:56:28 |
| 合計ジャッジ時間 | 1,308 ms |
|
ジャッジサーバーID (参考情報) |
judge4 / judge5 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 6 WA * 6 |
ソースコード
x, y, z = map(int, input().split())
s_line, s_num = input().split()
s_num = int(s_num)
e_line, e_num = input().split()
e_num = int(e_num)
if s_line == e_line:
print(abs(s_num - e_num))
else:
# All possible cases for different line transitions
# A to B
if s_line == 'A' and e_line == 'B':
cost = abs(s_num - 1) + 1 + abs(e_num - 1)
# A to C
elif s_line == 'A' and e_line == 'C':
cost1 = abs(s_num - 1) + 1 + (y - 1) + 1 + abs(z - e_num)
cost2 = abs(x - s_num) + 1 + abs(e_num - 1)
cost = min(cost1, cost2)
# B to A
elif s_line == 'B' and e_line == 'A':
cost1 = abs(s_num - 1) + 1 + abs(e_num - 1)
cost2 = abs(y - s_num) + 1 + abs(z - 1) + 1 + abs(x - e_num)
cost = min(cost1, cost2)
# B to C
elif s_line == 'B' and e_line == 'C':
cost = abs(y - s_num) + 1 + abs(z - e_num)
# C to A
elif s_line == 'C' and e_line == 'A':
cost1 = abs(s_num - 1) + 1 + abs(x - e_num)
cost2 = abs(z - s_num) + 1 + (y - 1) + 1 + abs(e_num - 1)
cost = min(cost1, cost2)
# C to B
elif s_line == 'C' and e_line == 'B':
cost = abs(z - s_num) + 1 + abs(y - e_num)
# B to C
elif s_line == 'B' and e_line == 'C':
cost = abs(y - s_num) + 1 + abs(z - e_num)
# C to B (duplicate case)
else:
# All other cases should be handled by above conditions
cost = 0 # This should theoretically never be reached
print(cost)
lam6er