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
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
このコードへのチャレンジ
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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
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 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()
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