結果

問題 No.622 点と三角柱の内外判定
ユーザー wajima_wataru
提出日時 2017-12-29 18:27:27
言語 Python3
(3.13.1 + numpy 2.2.1 + scipy 1.14.1)
結果
AC  
実行時間 39 ms / 1,500 ms
コード長 1,558 bytes
コンパイル時間 233 ms
コンパイル使用メモリ 12,928 KB
実行使用メモリ 11,264 KB
最終ジャッジ日時 2024-12-21 10:34:15
合計ジャッジ時間 2,573 ms
ジャッジサーバーID
(参考情報)
judge5 / judge1
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
other AC * 32
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

from math import sin, cos, atan2
from operator import *
pos = []
for _ in range(4):
pos.append(list(map(float, input().split())))
def rotate_x(p, theta):
q = []
q.append(p[0])
q.append(p[1] * cos(theta) - p[2] * sin(theta))
q.append(p[1] * sin(theta) + p[2] * cos(theta))
return q
def rotate_y(p, theta):
q = []
q.append(p[2] * sin(theta) + p[0] * cos(theta))
q.append(p[1])
q.append(p[2] * cos(theta) - p[0] * sin(theta))
return q
def rotate_z(p, theta):
q = []
q.append(p[0] * cos(theta) - p[1] * sin(theta))
q.append(p[0] * sin(theta) + p[1] * cos(theta))
q.append(p[2])
return q
def translate(p, q):
for i in range(3):
p[i] += q[i]
return p
d = list(map(neg, pos[0]))
for i in range(4):
pos[i] = translate(pos[i], d)
th1 = atan2(pos[1][2], pos[1][0])
for i in range(4):
pos[i] = rotate_y(pos[i], th1)
th2 = atan2(-pos[1][1], pos[1][0])
for i in range(4):
pos[i] = rotate_z(pos[i], th2)
th3 = atan2(-pos[2][2], pos[2][1])
for i in range(4):
pos[i] = rotate_x(pos[i], th3)
AD = [pos[3][0] - pos[0][0], pos[3][1] - pos[0][1]]
BD = [pos[3][0] - pos[1][0], pos[3][1] - pos[1][1]]
CD = [pos[3][0] - pos[2][0], pos[3][1] - pos[2][1]]
AB = [pos[1][0] - pos[0][0], pos[1][1] - pos[0][1]]
BC = [pos[2][0] - pos[1][0], pos[2][1] - pos[1][1]]
CA = [pos[0][0] - pos[2][0], pos[0][1] - pos[2][1]]
def cross(a, b):
return a[0] * b[1] - b[0] * a[1]
if cross(AB, BD) > 0 and cross(BC, CD) > 0 and cross(CA, AD) > 0:
print('YES')
elif cross(AB, BD) < 0 and cross(BC, CD) < 0 and cross(CA, AD) < 0:
print('YES')
else:
print('NO')
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