結果

問題 No.2180 Comprehensive Line Segments
ユーザー MasKoaTSMasKoaTS
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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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 305 ms
98,480 KB
testcase_01 MLE -
testcase_02 AC 127 ms
95,068 KB
testcase_03 MLE -
testcase_04 AC 164 ms
89,148 KB
testcase_05 AC 132 ms
88,392 KB
testcase_06 AC 127 ms
88,276 KB
testcase_07 AC 137 ms
88,396 KB
testcase_08 AC 134 ms
88,436 KB
testcase_09 AC 226 ms
98,824 KB
testcase_10 AC 326 ms
102,044 KB
testcase_11 MLE -
testcase_12 MLE -
testcase_13 TLE -
testcase_14 MLE -
testcase_15 MLE -
testcase_16 TLE -
testcase_17 MLE -
testcase_18 TLE -
testcase_19 MLE -
testcase_20 AC 232 ms
90,484 KB
testcase_21 MLE -
testcase_22 AC 403 ms
93,060 KB
testcase_23 AC 2,202 ms
220,756 KB
testcase_24 AC 887 ms
114,260 KB
testcase_25 TLE -
testcase_26 MLE -
testcase_27 WA -
testcase_28 MLE -
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ソースコード

diff # 

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)
0