結果

問題 No.2180 Comprehensive Line Segments
ユーザー MasKoaTS
提出日時 2022-10-13 12:06:15
言語 PyPy3
(7.3.15)
結果
TLE  
(最新)
AC  
(最初)
実行時間 -
コード長 1,627 bytes
コンパイル時間 216 ms
コンパイル使用メモリ 81,972 KB
実行使用メモリ 201,204 KB
最終ジャッジ日時 2024-11-17 01:20:51
合計ジャッジ時間 109,665 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge4
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 * 3 TLE * 1
other AC * 9 TLE * 16
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 itertools import permutations
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 __sub__(self, other):
        return Vector2(other.x - self.x, other.y - self.y)
    def __mul__(self, other):
        return self.x * other.y - self.y * other.x
def normalize_vector(v: Vector2) -> Vector2:
    assert(v.x != 0 or v.y != 0)
    norm = v.x ** 2 + v.y ** 2
    return Vector2(v.x * abs(v.x) / norm, v.y * abs(v.y) / norm)
def sgn(x: frac) -> int:
    if(x > 0):
        return 1
    if(x < 0):
        return -1
    return 0
"""
Main Code
"""
N = int(input())
P = [Vector2(*map(frac, input().split())) for _ in [0] * N]
if(N == 1):
    print(1)
    exit(0)
vec_lis = [[None]*N for _ in [0]*N]
for i in range(N - 1):
    for j in range(i + 1, N):
        vec_lis[i][j] = normalize_vector(P[j] - P[i])
        vec_lis[j][i] = normalize_vector(P[i] - P[j])
ans = N
num = [*range(N)]
for tup in permutations(num):
    p = [P[k] for k in num]
    lis = [vec_lis[tup[0]][tup[1]]]
    flag = False
    for i in range(1, N - 1):
        v = vec_lis[tup[i]][tup[i + 1]]
        if(lis[-1] == v):
            continue
        if(lis[-1] * v == 0):
            flag = True
            break
        lis.append(v)
    if(flag):
        continue
    M = len(lis)
    if(M <= 2):
        ans = min(ans, M)
        continue
    dp = [[-1]*2 for _ in [0]*(M - 1)]
    dp[0][0] = 0
    for j in range(1, M - 1):
        v1, v2, v3 = lis[j - 1: j + 2]
        if(sgn(v1 * v2) == sgn(v2 * v3) == sgn(v1 * v3)):
            dp[j][1] = dp[j - 1][0] + 1
        dp[j][0] = max(dp[j - 1])
    ans = min(ans, M - max(dp[M - 2]))
print(ans)
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