結果
| 問題 | No.1153 ねこちゃんゲーム |
| コンテスト | |
| ユーザー |
gew1fw
|
| 提出日時 | 2025-06-12 14:22:29 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 2,846 bytes |
| 記録 | |
| コンパイル時間 | 251 ms |
| コンパイル使用メモリ | 82,192 KB |
| 実行使用メモリ | 177,360 KB |
| 最終ジャッジ日時 | 2025-06-12 14:22:57 |
| 合計ジャッジ時間 | 27,322 ms |
|
ジャッジサーバーID (参考情報) |
judge3 / judge5 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 2 |
| other | AC * 6 WA * 29 TLE * 1 -- * 4 |
ソースコード
import sys
from sys import stdin
sys.setrecursionlimit(1 << 25)
def main():
import sys
input = sys.stdin.read
data = input().split()
idx = 0
N = int(data[idx])
idx += 1
M = int(data[idx])
idx += 1
A = list(map(int, data[idx:idx+M]))
idx += M
A = [a-1 for a in A]
# Read edges
edges = [[] for _ in range(N)]
for _ in range(N-1):
u = int(data[idx])-1
idx += 1
v = int(data[idx])-1
idx += 1
edges[u].append(v)
edges[v].append(u)
# Compute Grundy numbers. Root the tree at 0 (arbitrary)
from collections import deque
parent = [-1] * N
children = [[] for _ in range(N)]
visited = [False] * N
q = deque()
q.append(0)
visited[0] = True
while q:
u = q.popleft()
for v in edges[u]:
if not visited[v]:
visited[v] = True
parent[v] = u
children[u].append(v)
q.append(v)
# Compute Grundy numbers using post-order traversal
grundy = [0] * N
stack = [(0, False)]
while stack:
u, processed = stack.pop()
if processed:
mex = set()
for v in children[u]:
mex.add(grundy[v])
mex_val = 0
while mex_val in mex:
mex_val += 1
grundy[u] = mex_val
else:
stack.append((u, True))
for v in reversed(children[u]):
stack.append((v, False))
# Compute XOR of all cats' grundy numbers
xor = 0
for a in A:
xor ^= grundy[a]
if xor == 0:
print("-1 -1")
else:
# Find the best move for Ebi-chan
found = False
best_move = (-1, -1)
for i in range(M):
v = A[i]
for u in edges[v]:
if u == parent[v]:
continue # moving back to parent may not be optimal, but need to consider all
new_xor = xor ^ grundy[v] ^ grundy[u]
if new_xor == 0:
found = True
best_move = (i+1, u+1)
break
if found:
break
# If not found in children, check parent
if not found:
for i in range(M):
v = A[i]
if parent[v] != -1:
u = parent[v]
new_xor = xor ^ grundy[v] ^ grundy[u]
if new_xor == 0:
found = True
best_move = (i+1, u+1)
break
if found:
print(f"{best_move[0]} {best_move[1]}")
else:
# If no move found (unlikely as xor is non-zero)
print("1 -1")
if __name__ == '__main__':
main()
gew1fw