Python 找到给定点集的简单闭合路径（Find Simple Closed Path for a given set of points）

news2024/9/21 21:07:58

给定一组点，将这些点连接起来而不相交

例子：

输入：points[] = {(0, 3), (1, 1), (2, 2), (4, 4),
(0, 0), (1, 2), (3, 1}, {3, 3}};

输出：按以下顺序连接点将
不造成任何交叉
{(0, 0), (3, 1), (1, 1), (2, 2), (3, 3),
（4，4），（1，2），（0，3）}

我们强烈建议您最小化浏览器并先自己尝试一下。

这个想法是使用排序。

通过比较所有点的 y 坐标来找到最底部的点。如果有两个点的 y 值相同，则考虑 x 坐标值较小的点。将最底部的点放在第一个位置。

考虑剩余的 n-1 个点，并围绕 points[0] 按照极角逆时针顺序排列它们。如果两个点的极角相同，则将最近的点放在最前面。

遍历排序数组（按角度升序排序）产生简单的闭合路径。

如何计算角度？
一种解决方案是使用三角函数。
观察：我们不关心角度的实际值。我们只想按角度排序。
想法：使用方向来比较角度，而无需实际计算它们！

以下是上述想法的实现:

from functools import cmp_to_key
# A Python program to find simple closed path for n points
# for explanation of orientation()

# A global point needed for sorting points with reference
# to the first point. Used in compare function of qsort()
p0 = None

# A utility function to return square of distance between
# p1 and p2
def dist(p1, p2):
return (p1[0] - p2[0])*(p1[0] - p2[0]) + (p1[1] - p2[1])*(p1[1] - p2[1])

# To find orientation of ordered triplet (p, q, r).
# The function returns following values
# 0 --> p, q and r are collinear
# 1 --> Clockwise
# 2 --> Counterclockwise
def orientation(p, q, r):
val = (q[1] - p[1]) * (r[0] - q[0]) - (q[0] - p[0]) * (r[1] - q[1])

if val == 0: return 0 # collinear
return 1 if val > 0 else 2 # clockwise or counterclock wise

# A function used by library function qsort() to sort
# an array of points with respect to the first point
def compare(vp1, vp2):
p1 = vp1
p2 = vp2

# Find orientation
o = orientation(p0, p1, p2)
if o == 0:
return -1 if dist(p0, p2) >= dist(p0, p1) else 1

return -1 if o == 2 else 1

# Prints simple closed path for a set of n points.
def printClosedPath(points, n):
global p0
# Find the bottommost point
ymin = points[0][1]
min = 0
for i in range(1,n):
y = points[i][1]

# Pick the bottom-most. In case of tie, choose the
# left most point
if (y < ymin) or (ymin == y and points[i][0] < points[min][0]):
ymin = points[i][1]
min = i

# Place the bottom-most point at first position
temp = points[0]
points[0] = points[min]
points[min] = temp

# Sort n-1 points with respect to the first point.
# A point p1 comes before p2 in sorted output if p2
# has larger polar angle (in counterclockwise
# direction) than p1
p0 = points[0]
points.sort(key=cmp_to_key(compare))

# Now stack has the output points, print contents
# of stack
for i in range(n):
print("(",points[i][0],",",points[i][1],"), ", end="")

# Driver program to test above functions
points = [[0, 3], [1, 1], [2, 2], [4, 4], [0, 0], [1, 2], [3, 1], [3, 3]]
n = len(points)

printClosedPath(points, n)