結果

問題 No.2121 帰属関係と充足可能性
ユーザー lam6er
提出日時 2025-04-16 16:35:37
言語 PyPy3
(7.3.15)
結果
TLE  
実行時間 -
コード長 2,055 bytes
コンパイル時間 1,991 ms
コンパイル使用メモリ 81,676 KB
実行使用メモリ 110,832 KB
最終ジャッジ日時 2025-04-16 16:38:58
合計ジャッジ時間 4,299 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge1
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ファイルパターン 結果
other AC * 5 TLE * 1 -- * 43
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ソースコード

diff # 

import sys
from itertools import combinations

def generate_v(n):
    """Generate the cumulative hierarchy V_n as a list of frozensets."""
    if n == 0:
        return [frozenset()]
    prev = generate_v(n-1)
    subsets = set()
    for r in range(len(prev) + 1):
        for subset in combinations(prev, r):
            subsets.add(frozenset(subset))
    return list(subsets)

def main():
    input = sys.stdin.read().split()
    N = int(input[0])
    A = list(map(int, input[1:7]))
    A0, A1, A2, A3, A4, A5 = A

    # Generate V_N
    try:
        vn = generate_v(N)
    except RecursionError:
        print("NO")
        return

    # Precompute all elements of V_N
    elements = vn

    # Iterate all possible m0, m1, m2 in V_N
    found = False
    for m0 in elements:
        for m1 in elements:
            for m2 in elements:
                # Map A0-A5 to the corresponding m variables
                ma0 = [m0, m1, m2][A0]
                ma1 = [m0, m1, m2][A1]
                ma2 = [m0, m1, m2][A2]
                ma3 = [m0, m1, m2][A3]
                ma4 = [m0, m1, m2][A4]
                ma5 = [m0, m1, m2][A5]

                # Check condition 1: ∀x3 ∈ V_N, x3 ∈ ma0 → x3 ∈ ma1
                cond1 = True
                for x3 in elements:
                    if x3 in ma0:
                        if x3 not in ma1:
                            cond1 = False
                            break

                # Check condition 2: ma1 ∈ ma2
                cond2 = (ma1 in ma2)
                # Check condition 3: ma3 ∈ ma2
                cond3 = (ma3 in ma2)
                # Check condition 4: ma4 ∈ ma2
                cond4 = (ma4 in ma2)
                # Check condition 5: ma5 ∈ ma0
                cond5 = (ma5 in ma0)

                if cond1 and cond2 and cond3 and cond4 and cond5:
                    found = True
                    break
            if found:
                break
        if found:
            break

    print("YES" if found else "NO")

if __name__ == "__main__":
    main()
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