結果

問題 No.470 Inverse S+T Problem
コンテスト
ユーザー 👑 SPD_9X2
提出日時 2025-08-02 02:47:19
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 74 ms / 2,000 ms
コード長 4,472 bytes
コンパイル時間 336 ms
コンパイル使用メモリ 82,648 KB
実行使用メモリ 73,832 KB
最終ジャッジ日時 2025-08-02 02:47:24
合計ジャッジ時間 3,609 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge3
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 * 4
other AC * 27
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 # 

"""

2satかな

"""

import sys
import typing


class CSR:
    def __init__(
            self, n: int, edges: typing.List[typing.Tuple[int, int]]) -> None:
        self.start = [0] * (n + 1)
        self.elist = [0] * len(edges)

        for e in edges:
            self.start[e[0] + 1] += 1

        for i in range(1, n + 1):
            self.start[i] += self.start[i - 1]

        counter = self.start.copy()
        for e in edges:
            self.elist[counter[e[0]]] = e[1]
            counter[e[0]] += 1


class SCCGraph:
    '''
    Reference:
    R. Tarjan,
    Depth-First Search and Linear Graph Algorithms
    '''

    def __init__(self, n: int) -> None:
        self._n = n
        self._edges: typing.List[typing.Tuple[int, int]] = []

    def num_vertices(self) -> int:
        return self._n

    def add_edge(self, from_vertex: int, to_vertex: int) -> None:
        self._edges.append((from_vertex, to_vertex))

    def scc_ids(self) -> typing.Tuple[int, typing.List[int]]:
        g = CSR(self._n, self._edges)
        now_ord = 0
        group_num = 0
        visited = []
        low = [0] * self._n
        order = [-1] * self._n
        ids = [0] * self._n

        sys.setrecursionlimit(max(self._n + 1000, sys.getrecursionlimit()))

        def dfs(v: int) -> None:
            nonlocal now_ord
            nonlocal group_num
            nonlocal visited
            nonlocal low
            nonlocal order
            nonlocal ids

            low[v] = now_ord
            order[v] = now_ord
            now_ord += 1
            visited.append(v)
            for i in range(g.start[v], g.start[v + 1]):
                to = g.elist[i]
                if order[to] == -1:
                    dfs(to)
                    low[v] = min(low[v], low[to])
                else:
                    low[v] = min(low[v], order[to])

            if low[v] == order[v]:
                while True:
                    u = visited[-1]
                    visited.pop()
                    order[u] = self._n
                    ids[u] = group_num
                    if u == v:
                        break
                group_num += 1

        for i in range(self._n):
            if order[i] == -1:
                dfs(i)

        for i in range(self._n):
            ids[i] = group_num - 1 - ids[i]

        return group_num, ids

    def scc(self) -> typing.List[typing.List[int]]:
        ids = self.scc_ids()
        group_num = ids[0]
        counts = [0] * group_num
        for x in ids[1]:
            counts[x] += 1
        groups: typing.List[typing.List[int]] = [[] for _ in range(group_num)]
        for i in range(self._n):
            groups[ids[1][i]].append(i)

        return groups


class TwoSAT:
    '''
    2-SAT

    Reference:
    B. Aspvall, M. Plass, and R. Tarjan,
    A Linear-Time Algorithm for Testing the Truth of Certain Quantified Boolean
    Formulas
    '''

    def __init__(self, n: int = 0) -> None:
        self._n = n
        self._answer = [False] * n
        self._scc = SCCGraph(2 * n)

    def add_clause(self, i: int, f: bool, j: int, g: bool) -> None:
        assert 0 <= i < self._n
        assert 0 <= j < self._n

        self._scc.add_edge(2 * i + (0 if f else 1), 2 * j + (1 if g else 0))
        self._scc.add_edge(2 * j + (0 if g else 1), 2 * i + (1 if f else 0))

    def satisfiable(self) -> bool:
        scc_id = self._scc.scc_ids()[1]
        for i in range(self._n):
            if scc_id[2 * i] == scc_id[2 * i + 1]:
                return False
            self._answer[i] = scc_id[2 * i] < scc_id[2 * i + 1]
        return True

    def answer(self) -> typing.List[bool]:
        return self._answer

N = int(input())

if N > 26*2:
    print ("Impossible")
    sys.exit()

ts = TwoSAT(N)

U = [ input() for i in range(N) ]

for i in range(N):
    for j in range(i+1,N):

        a = U[i]
        b = U[j]
        
        #0 vs 0
        if a[0] == b[0] or a[1:] == b[1:]:
            ts.add_clause(i,1,j,1)

        #0vs1
        if a[0] == b[2] or a[1:] == b[:2]:
            ts.add_clause(i,1,j,0)

        #1vs0
        if a[2] == b[0] or a[:2] == b[1:]:
            ts.add_clause(i,0,j,1)

        #1vs1
        if a[2] == b[2] or a[:2] == b[:2]:
            ts.add_clause(i,0,j,0)

if not ts.satisfiable():
    print ("Impossible")
else:
    ans = ts.answer()

    for i,s in enumerate(U):
        if ans[i]:
            print (s[:2],s[2])
        else:
            print (s[0],s[1:])
0