結果

問題 No.3031 曲面の向き付け
コンテスト
ユーザー kidodesu
提出日時 2025-02-21 23:09:42
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 1,956 ms / 2,000 ms
コード長 5,185 bytes
コンパイル時間 294 ms
コンパイル使用メモリ 82,352 KB
実行使用メモリ 425,220 KB
最終ジャッジ日時 2025-02-21 23:09:56
合計ジャッジ時間 13,571 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge3
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 * 4
other AC * 29
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
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())
tri = [tuple(map(int, input().split())) for _ in range(N)]
from collections import defaultdict
dict = defaultdict(lambda: -1)

sat = TwoSAT(N)

def make(a, b):
    return (a << 30) | b

for i in range(N):
    a, b, c = tri[i]
    k0 = make(a, b)
    k1 = make(b, c)
    k2 = make(a, c)
    if dict[k0] == -1:
        dict[k0] = i*2
    elif dict[k0] == -2:
        exit(print("NO"))
    else:
        x = dict[k0]
        j = x // 2
        if x % 2:
            sat.add_clause(i, True, j, False)
            sat.add_clause(i, False, j, True)
        else:
            sat.add_clause(i, True, j, True)
            sat.add_clause(i, False, j, False)
        dict[k0] = -2
    if dict[k1] == -1:
        dict[k1] = i*2
    elif dict[k1] == -2:
        exit(print("NO"))
    else:
        x = dict[k1]
        j = x // 2
        if x % 2:
            sat.add_clause(i, True, j, False)
            sat.add_clause(i, False, j, True)
        else:
            sat.add_clause(i, True, j, True)
            sat.add_clause(i, False, j, False)
        dict[k1] = -2
    if dict[k2] == -1:
        dict[k2] = i*2+1
    elif dict[k2] == -2:
        exit(print("NO"))
    else:
        x = dict[k2]
        j = x // 2
        if not (x % 2):
            sat.add_clause(i, True, j, False)
            sat.add_clause(i, False, j, True)
        else:
            sat.add_clause(i, True, j, True)
            sat.add_clause(i, False, j, False)
        dict[k2] = -2

if sat.satisfiable():
    print("YES")
else:
    print("NO")
0