【计算几何】Python求解覆盖N个点的最小圆
题目地址
https://ac.nowcoder.com/acm/contest/52826/D
代码
import sys
import math
def euclidean_distance(p1, p2):
return math.sqrt((p1[0] - p2[0]) ** 2 + (p1[1] - p2[1]) ** 2)
def make_circle(points):
shuffled = list(points)
circle = (0, 0, 0)
for i, p in enumerate(shuffled):
if is_inside_circle(circle, p):
continue
circle = (p[0], p[1], 0)
for j, q in enumerate(shuffled[:i]):
if is_inside_circle(circle, q):
continue
circle = ((p[0] + q[0]) / 2, (p[1] + q[1]) / 2, euclidean_distance(p, q) / 2)
for k, r in enumerate(shuffled[:j]):
if is_inside_circle(circle, r):
continue
circle = make_circumcircle(p, q, r)
return circle
def is_inside_circle(circle, point):
center, radius = (circle[0], circle[1]), circle[2]
return euclidean_distance(center, point) <= radius
def make_circumcircle(p1, p2, p3):
d = 2 * (p1[0] * (p2[1] - p3[1]) + p2[0] * (p3[1] - p1[1]) + p3[0] * (p1[1] - p2[1]))
ux = ((p1[0] ** 2 + p1[1] ** 2) * (p2[1] - p3[1]) + (p2[0] ** 2 + p2[1] ** 2) * (p3[1] - p1[1]) + (p3[0] ** 2 + p3[1] ** 2) * (p1[1] - p2[1])) / d
uy = ((p1[0] ** 2 + p1[1] ** 2) * (p3[0] - p2[0]) + (p2[0] ** 2 + p2[1] ** 2) * (p1[0] - p3[0]) + (p3[0] ** 2 + p3[1] ** 2) * (p2[0] - p1[0])) / d
r = euclidean_distance((ux, uy), p1)
return (ux, uy, r)
def main():
N = int(input().strip())
points = [tuple(map(float, input().strip().split())) for _ in range(N)]
circle = make_circle(points)
print("{:.10f}".format(circle[2]))
print("{:.10f} {:.10f}".format(circle[0], circle[1]))
if __name__ == "__main__":
main()
Prompt
Given N points, let's draw the smallest circle that contains all the points.
python code
input
6
8.0 9.0
4.0 7.5
1.0 2.0
5.1 8.7
9.0 2.0
4.5 1.0
output
5.0000000000
5.0000000000 5.0000000000
