結果
| 問題 |
No.408 五輪ピック
|
| コンテスト | |
| ユーザー |
gew1fw
|
| 提出日時 | 2025-06-12 13:54:45 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
MLE
|
| 実行時間 | - |
| コード長 | 1,408 bytes |
| コンパイル時間 | 378 ms |
| コンパイル使用メモリ | 82,632 KB |
| 実行使用メモリ | 204,420 KB |
| 最終ジャッジ日時 | 2025-06-12 13:56:07 |
| 合計ジャッジ時間 | 8,895 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 |
|---|---|
| other | AC * 23 MLE * 5 -- * 4 |
ソースコード
from sys import stdin
def main():
input = stdin.read().split()
idx = 0
n = int(input[idx])
idx += 1
m = int(input[idx])
idx += 1
adj = [[] for _ in range(n + 1)]
for _ in range(m):
u = int(input[idx])
idx += 1
v = int(input[idx])
idx += 1
adj[u].append(v)
adj[v].append(u)
# Collect neighbors of vertex 1
s = set()
for neighbor in adj[1]:
s.add(neighbor)
if not s:
print("NO")
return
found = False
# Check each neighbor of 1
for a in s:
# DFS stack: (current node, depth, visited set)
stack = [(a, 0, {1, a})]
while stack:
node, depth, visited = stack.pop()
if depth == 3:
if node in s and node != a:
found = True
break
continue
# Explore neighbors in reverse order to process smaller indices first (stack order)
for neighbor in reversed(adj[node]):
if neighbor == 1:
continue
if neighbor not in visited:
new_visited = visited.copy()
new_visited.add(neighbor)
stack.append((neighbor, depth + 1, new_visited))
if found:
break
print("YES" if found else "NO")
if __name__ == "__main__":
main()
gew1fw