結果

問題 No.2180 Comprehensive Line Segments
ユーザー MasKoaTS
提出日時 2022-11-28 17:36:35
言語 PyPy3
(7.3.15)
結果
MLE  
(最新)
AC  
(最初)
実行時間 -
コード長 3,459 bytes
コンパイル時間 276 ms
コンパイル使用メモリ 82,716 KB
実行使用メモリ 1,622,432 KB
最終ジャッジ日時 2024-11-17 01:41:05
合計ジャッジ時間 67,738 ms
ジャッジサーバーID
（参考情報） 		judge5 / 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 * 2 MLE * 2
other AC * 11 WA * 1 TLE * 4 MLE * 9
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 #
プレゼンテーションモードにする

import sys
from collections import deque
from fractions import Fraction as frac
input = sys.stdin.readline
class Vector2:
    def __init__(self, x: frac, y: frac):
        self.x = x
        self.y = y
    def __eq__(self, other):
        return (self.x == other.x and self.y == other.y)
    def __hash__(self):
        return hash((self.x, self.y))
    def __sub__(self, other):
        return Vector2(other.x - self.x, other.y - self.y)
    def normalize(self):
        assert(self.x != 0 or self.y != 0)
        norm = self.x ** 2 + self.y ** 2
        self.x *= abs(self.x) / norm
        self.y *= abs(self.y) / norm
        return self
class Line:
    def __init__(self, a: frac, b:frac, c:frac):
        self.a = a
        self.b = b
        self.c = c
    def __eq__(self, other):
        return (self.a == other.a and self.b == other.b and self.c == other.c)
    def __hash__(self):
        return hash((self.a, self.b, self.c))
def calc_line(one: Vector2, other: Vector2) -> Line:
    if(one == other):
        return None
    x1 = one.x;  y1 = one.y
    x2 = other.x;  y2 = other.y
    if(x1 == x2):
        return Line(1, 0, x1)
    a = (y1 - y2) / (x1 - x2)
    c = y1 - a * x1
    return Line(-a, 1, c)
def calc_intersection(one: Line, other: Line) -> Vector2:
    p = one.a * other.b - other.a * one.b
    if(p == 0):
        return None
    q = other.b * one.c - one.b * other.c
    x = q / p
    y = (other.c - other.a * x) / other.b if(one.b == 0) else (one.c - one.a * x) / one.b
    return Vector2(x, y)
"""
Main Code
"""
# 入力
N = int(input())
P = [Vector2(*map(frac, input().split())) for _ in [0] * N]
# 点が 1 個のときは必ず答えは 1
if(N == 1):
    print(1)
    exit(0)
# 各点に番号付け
point_dic = {}
point_num = 0
for p in P:
    point_dic[p] = point_num
    point_num += 1
# 有り得る直線を調べて番号付け
line_dic = {}
line_num = 0
for i in range(N - 1):
    for j in range(i + 1, N):
        p1, p2 = P[i], P[j]
        line = calc_line(p1, p2)
        if(line in line_dic):
            continue
        line_dic[line] = line_num
        line_num += 1
# 有り得る交点を調べて番号付け
line_lis = list(line_dic.keys())
for i in range(line_num - 1):
    for j in range(i + 1, line_num):
        l1, l2 = line_lis[i], line_lis[j]
        p = calc_intersection(l1, l2)
        if((p is None) or p in point_dic):
            continue
        P.append(p)
        point_dic[p] = point_num
        point_num += 1
# 任意の2点について、2点から構成されるベクトルを調べて正規化
vec_lis = [[None]*point_num for _ in [0]*point_num]
for i in range(point_num - 1):
    for j in range(i + 1, point_num):
        if(calc_line(P[i], P[j]) not in line_dic):
            continue
        vec_lis[i][j] = (P[j] - P[i]).normalize()
        vec_lis[j][i] = (P[i] - P[j]).normalize()
# グラフ探索
dp = [[[N]*point_num for _ in [0]*point_num] for _ in [0]*(1 << N)]
que = deque([])
for i in range(N):
    que.append((0, 1 << i, i, i))
    dp[1 << i][i][i] = 0
goal = (1 << N) - 1
ans = N
while(que):
    c_now, b_now, v_prev, v_now = que.popleft()
    if(c_now > dp[b_now][v_prev][v_now]):
        continue
    if(b_now == goal):
        ans = c_now
        break
    for v_next in range(point_num):
        if(v_now == v_next or min(v_now, v_next) >= N or (vec_lis[v_now][v_next] is None)):
            continue
        b_next = b_now | (1 << v_next) if(v_next < N) else b_now
        c_next = c_now
        if(v_prev == v_now or vec_lis[v_prev][v_now] != vec_lis[v_now][v_next]):
            c_next += 1
        if(c_next >= dp[b_next][v_now][v_next]):
            continue
        dp[b_next][v_now][v_next] = c_next
        if(c_next == c_now):
            que.appendleft((c_next, b_next, v_now, v_next))
        else:
            que.append((c_next, b_next, v_now, v_next))
print(ans)
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