結果

問題 No.2180 Comprehensive Line Segments
ユーザー MasKoaTSMasKoaTS
提出日時 2022-11-30 21:42:59
言語 PyPy3
(7.3.15)
結果
MLE  
(最新)
AC  
(最初)
実行時間 -
コード長 3,481 bytes
コンパイル時間 150 ms
コンパイル使用メモリ 82,056 KB
実行使用メモリ 849,800 KB
最終ジャッジ日時 2024-04-28 10:36:04
合計ジャッジ時間 6,297 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge4
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 301 ms
91,948 KB
testcase_01 AC 131 ms
88,400 KB
testcase_02 AC 127 ms
88,316 KB
testcase_03 MLE -
testcase_04 -- -
testcase_05 -- -
testcase_06 -- -
testcase_07 -- -
testcase_08 -- -
testcase_09 -- -
testcase_10 -- -
testcase_11 -- -
testcase_12 -- -
testcase_13 -- -
testcase_14 -- -
testcase_15 -- -
testcase_16 -- -
testcase_17 -- -
testcase_18 -- -
testcase_19 -- -
testcase_20 -- -
testcase_21 -- -
testcase_22 -- -
testcase_23 -- -
testcase_24 -- -
testcase_25 -- -
testcase_26 -- -
testcase_27 -- -
testcase_28 -- -
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 # 

from __future__ import annotations
import sys
from collections import deque
from itertools import combinations
from fractions import Fraction as frac
input = sys.stdin.readline


class Vector2:
	def __init__(self, x, y):
		self.x = x
		self.y = y

	def __str__(self):
		return str((self.x, self.y))

	__repr__ = __str__

	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) -> Vector2:
		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, b, c):
		self.a = a
		self.b = b
		self.c = c

	def __str__(self):
		return str((self.a, self.b, self.c))

	__repr__ = __str__

	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))
	
	@classmethod
	def calc_line(cls, 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)

	@classmethod
	def calc_intersection(cls, 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)

# 任意の 2 点に関する直線を調べてsetに追加
line_set = set([])
for p1, p2 in combinations(P, 2):
	line_set.add(Line.calc_line(p1, p2))
	
# 任意の 2 直線の交点を調べてlist, setに追加
point_set = set(P)
point_num = N
for l1, l2 in combinations(line_set, 2):
	p = Line.calc_intersection(l1, l2)
	if(p is None or p in point_set):
		continue
	point_set.add(p)
	P.append(p)
	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(Line.calc_line(P[i], P[j]) not in line_set):
			continue
		vec_lis[i][j] = (P[i] - P[j]).normalize()
		vec_lis[j][i] = (P[j] - P[i]).normalize()

# グラフ探索
mbit = 1 << N
goal = mbit - 1
dp = [[[N]*point_num for _ in [0]*point_num] for _ in [0]*mbit]
que = deque([])
ans = N
for i in range(N):
	dp[1 << i][i][i] = 0
	que.append((1 << i, i, i, 0))
while(que):
	b_now, v_prev, v_now, c_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_next in [v_prev, v_now] or min(v_now, v_next) >= N):
			continue
		if(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 + (v_prev == v_now or vec_lis[v_prev][v_now] != vec_lis[v_now][v_next])
		if(c_next >= dp[b_next][v_now][v_next]):
			continue
		dp[b_next][v_now][v_next] = c_next
		if(c_now == c_next):
			que.appendleft((b_next, v_now, v_next, c_next))
		else:
			que.append((b_next, v_now, v_next, c_next))

print(ans)
0