你能不能用Python帮我写一个计算多个矩形重叠形成的多边形边长的程序

可以的，以下是一个计算多个矩形重叠形成的多边形边长的Python程序：

import numpy as np

# 输入矩形的坐标和数量
n = int(input("请输入矩形的数量："))
rects = []
for i in range(n):
    x1, y1, x2, y2 = map(int, input(f"请输入第{i+1}个矩形的坐标：").split())
    rects.append([(x1, y1), (x2, y1), (x2, y2), (x1, y2)])

# 计算矩形重叠部分的多边形
polygons = []
for i in range(n):
    polygon = rects[i]
    for j in range(i+1, n):
        rect = rects[j]
        intersection = []
        for k in range(4):
            xk, yk = polygon[k]
            xk1, yk1 = polygon[(k+1)%4]
            for l in range(4):
                xl, yl = rect[l]
                xl1, yl1 = rect[(l+1)%4]
                if (xk-xk1)*(yl-yl1) == (yk-yk1)*(xl-xl1):
                    if min(xk, xk1) <= xl <= max(xk, xk1) and min(yl, yl1) <= yk <= max(yl, yl1):
                        intersection.append((xl, yk))
                    elif min(xk, xk1) <= xl1 <= max(xk, xk1) and min(yl, yl1) <= yk <= max(yl, yl1):
                        intersection.append((xl1, yk))
                    elif min(xk, xk1) <= xk <= max(xk, xk1) and min(yl, yl1) <= yl <= max(yl, yl1):
                        intersection.append((xk, yl))
                    elif min(xk, xk1) <= xk1 <= max(xk, xk1) and min(yl, yl1) <= yl <= max(yl, yl1):
                        intersection.append((xk1, yl))
        if intersection:
            intersection = np.array(intersection)
            hull = np.concatenate([polygon, rect, intersection])
            hull = np.unique(hull, axis=0)
            hull = [tuple(point) for point in hull]
            polygons.append(hull)
    polygons.append(polygon)

# 计算多边形的边长
perimeters = []
for polygon in polygons:
    perimeter = 0
    for i in range(len(polygon)):
        x1, y1 = polygon[i]
        x2, y2 = polygon[(i+1)%len(polygon)]
        perimeter += ((x2-x1)**2 + (y2-y1)**2)**0.5
    perimeters.append(perimeter)

# 输出结果
for i in range(len(polygons)):
    print(f"第{i+1}个多边形的边长为：{perimeters[i]}")

该程序首先输入矩形的坐标和数量，然后计算矩形重叠部分的多边形，并计算多边形的边长，最后输出每个多边形的边长。程序中使用了NumPy库来处理向量和矩阵运算。