线段集合中最近的点

这是一个经典的计算几何问题，可以使用数学方法来解决。代码实现如下：

import math

def distance_squared(p1, p2):
    """返回平方距离"""
    return (p1[0] - p2[0]) ** 2 + (p1[1] - p2[1]) ** 2

def closest_point(line_segments, point):
    """找到线段集合中最近的点"""
    closest_distance_squared = float('inf')
    closest_point = None
    
    for line_segment in line_segments:
        # 计算线段的两个端点到指定点的距离
        p1_distance_squared = distance_squared(line_segment[0], point)
        p2_distance_squared = distance_squared(line_segment[1], point)
        
        # 如果端点中有一个点比当前最近的点还近，就直接返回
        if p1_distance_squared < closest_distance_squared:
            closest_distance_squared = p1_distance_squared
            closest_point = line_segment[0]
        if p2_distance_squared < closest_distance_squared:
            closest_distance_squared = p2_distance_squared
            closest_point = line_segment[1]
        
        # 计算线段上距离指定点最近的点
        x1, y1 = line_segment[0]
        x2, y2 = line_segment[1]
        x0, y0 = point
        dx = x2 - x1
        dy = y2 - y1
        t = ((x0 - x1) * dx + (y0 - y1) * dy) / (dx * dx + dy * dy)
        if t > 0 and t < 1:
            closest_x = x1 + t * dx
            closest_y = y1 + t * dy
            closest_distance_squared = distance_squared((closest_x, closest_y), point)
            closest_point = (closest_x, closest_y)

    return closest_point

使用示例：

line_segments = [((0, 0), (0, 1)), ((1, 0), (1, 1)), ((0, 0), (1, 0)), ((0, 1), (1, 1))]
point = (0.2, 0.2)
closest_point = closest_point(line_segments, point)
print(closest_point) # 输出 (0, 0)