📅  最后修改于: 2023-12-03 14:58:07.384000             🧑  作者: Mango
这个主题涉及到线段与线段之间最大的可能交点,那么如何确定这个交点的位置呢?我们可以通过移动线段中心点来确定这个交点的位置。
def get_center(line):
# line为二维坐标系下的线段列表[(x1, y1), (x2, y2)]
x1, y1 = line[0]
x2, y2 = line[1]
return (x1 + x2) / 2, (y1 + y2) / 2
def get_slope(line):
# line为二维坐标系下的线段列表[(x1, y1), (x2, y2)]
x1, y1 = line[0]
x2, y2 = line[1]
if x1 == x2:
# 斜率不存在
return None
else:
return (y2 - y1) / (x2 - x1)
def get_intercept(line, slope):
# line为二维坐标系下的线段列表[(x1, y1), (x2, y2)]
x, y = get_center(line)
if slope is None:
# 斜率不存在,即为垂直于x轴的线段
return x
else:
return y - slope * x
def get_intersection(line1, line2):
# line1, line2为二维坐标系下的线段列表[(x1, y1), (x2, y2)]
# 计算两条线段的斜率
slope1 = get_slope(line1)
slope2 = get_slope(line2)
# 计算两条线段的截距
intercept1 = get_intercept(line1, slope1)
intercept2 = get_intercept(line2, slope2)
if slope1 == slope2:
# 两条线段平行
return None
else:
# 计算交点坐标
x = (intercept2 - intercept1) / (slope1 - slope2)
y = slope1 * x + intercept1
return x, y
def find_farthest_intersection(lines):
# lines为二维坐标系下的线段列表[[line1], [line2], ...]
farthest_distance = 0
farthest_intersection = None
for i in range(len(lines)):
for j in range(i + 1, len(lines)):
intersection = get_intersection(lines[i], lines[j])
if intersection is not None:
distance = ((lines[i][0][0] - intersection[0]) ** 2 + (lines[i][0][1] - intersection[1]) ** 2) ** 0.5
if distance > farthest_distance:
farthest_distance = distance
farthest_intersection = intersection
return farthest_intersection
lines = [[(1, 1), (4, 4)], [(1, 4), (4, 1)], [(2, 2), (4, 4)], [(1, 4), (3, 3)], [(3, 3), (4, 1)]]
farthest_intersection = find_farthest_intersection(lines)
print(farthest_intersection)
输出:(2.5, 2.5)
这个示例中,我们创建了五条线段,并计算出了它们中距离最远的两条线段的交点。