結果

問題 No.2632 Center of Three Points in Lp Norm
ユーザー tatyamtatyam
提出日時 2024-02-16 22:38:46
言語 Python3
(3.13.1 + numpy 2.2.1 + scipy 1.14.1)
結果
AC  
実行時間 589 ms / 2,000 ms
コード長 1,628 bytes
コンパイル時間 409 ms
コンパイル使用メモリ 12,928 KB
実行使用メモリ 44,348 KB
最終ジャッジ日時 2024-09-28 21:15:03
合計ジャッジ時間 24,704 ms
ジャッジサーバーID
(参考情報)
judge3 / judge5
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 * 4
other AC * 34
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 numpy as np
from math import pi, cos, sin, atan2
import sys
ZERO = np.array([0.0, 0.0])
get = lambda: np.array(list(map(float, input().split())))
def norm(a) -> float:
x = abs(a[0])
y = abs(a[1])
if x > y:
x, y = y, x
if y == 0.0:
return 0.0
x /= y
return pow(1.0 + pow(x, p), 1 / p) * y
def dist(a, b):
return norm(a - b)
def cross(a, b):
return a[0] * b[1] - a[1] * b[0]
p = float(input())
a = get()
b = get()
c = get()
if a[0] > b[0]:
a, b = b, a
if a[0] > c[0]:
a, c = c, a
if cross(b - a, c - a) < 0:
b, c = c, b
A = a.copy()
a -= A
b -= A
c -= A
B = -pi
C = pi
def atan(diff) -> float:
return atan2(diff[1], diff[0])
def near(diff) -> int:
da = dist(a, diff)
db = dist(b, diff)
dc = dist(c, diff)
mn = min(da, db, dc)
if da == mn: return 0
if db == mn: return 1
return 2
def bisect_a(di):
ok = 1e7
ng = 0.0
for _ in range(100):
mid = (ok + ng) / 2
if near(di * mid) == 0:
ng = mid
else:
ok = mid
return di * ok
def direction(theta):
return np.array([cos(theta), sin(theta)])
def check(theta):
di = direction(theta)
x = bisect_a(di)
n = near(x)
if n:
return n
if theta < atan(b):
return 1
return 2
for _ in range(100):
theta = (B + C) / 2
if check(theta) == 1:
B = theta
else:
C = theta
ans = bisect_a(direction(B))
if max(abs(ans[0]), abs(ans[1])) > 1e6:
ans = bisect_a(direction(C))
print(dist(a, ans), dist(b, ans), dist(c, ans), file=sys.stderr)
print(*(ans + A))
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