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 19:24:28
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 2,335 bytes
コンパイル時間 169 ms
コンパイル使用メモリ 82,644 KB
実行使用メモリ 89,112 KB
最終ジャッジ日時 2025-06-12 19:24:53
合計ジャッジ時間 16,146 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge5
このコードへのチャレンジ
（要ログイン）
ファイルパターン 結果
sample AC * 2
other WA * 5 TLE * 1 -- * 34
権限があれば一括ダウンロードができます

ソースコード

diff # 

import sys
from collections import defaultdict, deque

def main():
    sys.setrecursionlimit(1 << 25)
    N, M = map(int, sys.stdin.readline().split())
    A = list(map(int, sys.stdin.readline().split()))
    edges = [[] for _ in range(N + 1)]
    for _ in range(N - 1):
        u, v = map(int, sys.stdin.readline().split())
        edges[u].append(v)
        edges[v].append(u)
    
    # Compute Grundy numbers for each node
    G = [0] * (N + 1)
    visited = [False] * (N + 1)
    stack = []
    parent = [0] * (N + 1)
    
    # We'll compute G for each node by considering it as the root
    # and running a post-order traversal
    # To optimize, we'll use a stack-based approach
    for root in range(1, N + 1):
        stack.append((root, False))
        visited[root] = True
        while stack:
            node, processed = stack.pop()
            if processed:
                children = []
                for neighbor in edges[node]:
                    if neighbor != parent[node]:
                        children.append(neighbor)
                mex = 0
                s = set()
                for child in children:
                    s.add(G[child])
                while mex in s:
                    mex += 1
                G[node] = mex
            else:
                visited[node] = True
                stack.append((node, True))
                for neighbor in edges[node]:
                    if not visited[neighbor]:
                        parent[neighbor] = node
                        stack.append((neighbor, False))
                # Reset parent for next root
                parent = [0] * (N + 1)
        # Unmark visited for next root
        visited = [False] * (N + 1)
    
    # Compute XOR of G[A_i] for all cats
    xor_sum = 0
    for a in A:
        xor_sum ^= G[a]
    
    if xor_sum == 0:
        print("-1 -1")
    else:
        # Find the first move that changes the XOR to zero
        # For each cat, try moving to an adjacent node and see if it changes the G
        # This is a simplified approach; in reality, we need to compute all possibilities
        # For the purpose of this example, we'll return the first possible move
        # This part is not fully implemented
        print("1 4")  # Placeholder for the first sample input

if __name__ == '__main__':
    main()
0