結果

問題 No.2180 Comprehensive Line Segments
ユーザー MasKoaTSMasKoaTS
提出日時 2023-08-13 12:14:17
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 1,987 ms / 5,000 ms
コード長 3,415 bytes
コンパイル時間 321 ms
コンパイル使用メモリ 82,560 KB
実行使用メモリ 90,164 KB
最終ジャッジ日時 2024-11-21 06:54:34
合計ジャッジ時間 20,754 ms
ジャッジサーバーID
（参考情報） 		judge2 / 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 96 ms
76,648 KB
testcase_01 AC 45 ms
53,760 KB
testcase_02 AC 39 ms
53,120 KB
testcase_03 AC 1,790 ms
87,760 KB
testcase_04 AC 42 ms
54,144 KB
testcase_05 AC 40 ms
53,376 KB
testcase_06 AC 39 ms
53,632 KB
testcase_07 AC 39 ms
53,376 KB
testcase_08 AC 40 ms
53,632 KB
testcase_09 AC 1,111 ms
86,576 KB
testcase_10 AC 1,145 ms
88,908 KB
testcase_11 AC 1,909 ms
89,832 KB
testcase_12 AC 664 ms
81,992 KB
testcase_13 AC 1,840 ms
88,156 KB
testcase_14 AC 1,645 ms
90,148 KB
testcase_15 AC 1,987 ms
88,900 KB
testcase_16 AC 285 ms
80,172 KB
testcase_17 AC 349 ms
79,608 KB
testcase_18 AC 1,824 ms
88,760 KB
testcase_19 AC 665 ms
81,788 KB
testcase_20 AC 51 ms
62,976 KB
testcase_21 AC 379 ms
81,116 KB
testcase_22 AC 95 ms
76,416 KB
testcase_23 AC 205 ms
78,428 KB
testcase_24 AC 157 ms
78,168 KB
testcase_25 AC 244 ms
79,376 KB
testcase_26 AC 1,803 ms
90,164 KB
testcase_27 AC 528 ms
81,816 KB
testcase_28 AC 523 ms
82,996 KB
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 itertools import permutations
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 __neg__(self):
        return Vector2(-self.x, -self.y)

    def __add__(self, other):
        return Vector2(self.x + other.x, self.y + other.y)

    def __sub__(self, other):
        return Vector2(self.x - other.x, self.y - other.y)

    def __mul__(self, other):
        return self.x * other.y - self.y * other.x

    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


def sgn(x: int) -> int:
    if(x > 0):
        return 1
    if(x < 0):
        return -1
    return 0

def same_inclination(v1: Vector2, v2: Vector2) -> bool:
    return (v1 * v2 == 0 and (v1.x * v2.x > 0 or v1.y * v2.y > 0))


"""
Main Code
"""

N = int(input())
P = [Vector2(*map(int, input().split())) for _ in [0] * N]

if(N == 1):
    print(1)
    exit(0)

perm = [*permutations([*range(N)], 2)]
vectors = [[None]*N for _ in [0]*N]
for i, j in perm:
    vectors[i][j] = P[j] - P[i]

ans = N
dp = [[[N]*N for _ in [0]*N] for _ in [0]*(1 << N)]
for i, j in perm:
    dp[(1 << i) | (1 << j)][i][j] = 1

goal = (1 << N) - 1
for b_now in range(3, goal):
    for v_prev, v_now in perm:
        if(dp[b_now][v_prev][v_now] == N or not((b_now >> v_prev) & 1) or not((b_now >> v_now) & 1)):
            continue
        c_next = c_now = dp[b_now][v_prev][v_now]
        v1 = vectors[v_prev][v_now]
        for v_next1, v_next2 in perm:
            if((((b_now >> v_next1) & 1) and v_now != v_next1) or ((b_now >> v_next2) & 1)):
                continue
            b_next = b_now | (1 << v_next1) | (1 << v_next2)
            v2, v3 = vectors[v_now][v_next1], vectors[v_next1][v_next2]
            d = e = 1
            if(v_now == v_next1):
                c_next = c_now + 1 - same_inclination(v1, v3)
            else:
                d = e = same_inclination(v1, v2) + same_inclination(v2, v3)
                if(e == 0):
                    e = int(sgn(v1 * v2) == sgn(v2 * v3) == sgn(v1 * v3))
                c_next = c_now + 2 - e
            if(dp[b_next][v_next1][v_next2] > c_next):
                dp[b_next][v_next1][v_next2] = c_next
                
            # 1点だけ通る場合を考慮
            if(v_now == v_next1 or d != 0):
                continue
            c_in = c_next + e
            for v_in in range(N):
                if((b_next >> v_in) & 1):
                    continue
                v4 = vectors[v_now][v_in]
                v5 = vectors[v_next1][v_in]
                if not(sgn(v1 * v4) == sgn(v4 * v2) == sgn(v1 * v2)):
                    continue
                if not(sgn((-v2) * v5) == sgn(v5 * (-v3)) == sgn((-v2) * (-v3))):
                    continue
                b_in = b_next | (1 << v_in)
                if(dp[b_in][v_next1][v_next2] <= c_in):
                    continue
                dp[b_in][v_next1][v_next2] = c_in

ans = min(min(lis) for lis in dp[goal])
print(ans)
0