三角剖分python代码
时间: 2023-10-13 16:23:28 浏览: 158
以下是一个基于Delaunay三角剖分的Python代码示例:
```python
import numpy as np
import matplotlib.pyplot as plt
from scipy.spatial import Delaunay
# 创建一组随机的点
points = np.random.rand(30, 2)
# 进行Delaunay三角剖分
tri = Delaunay(points)
# 绘制三角剖分结果
plt.triplot(points[:,0], points[:,1], tri.simplices)
plt.plot(points[:,0], points[:,1], 'o')
plt.show()
```
代码中,首先使用numpy库生成30个随机点,然后使用scipy库中的Delaunay函数进行三角剖分,并使用matplotlib库进行绘图展示。
相关问题
凸多边形最优三角剖分python代码和伪代码
以下是凸多边形最优三角剖分的 Python 代码和伪代码:
Python 代码:
```python
import math
# 计算三角形面积
def triangle_area(p1, p2, p3):
return abs(0.5 * ((p2[0] - p1[0]) * (p3[1] - p1[1]) - (p2[1] - p1[1]) * (p3[0] - p1[0])))
# 计算最优三角剖分
def optimal_triangulation(points):
n = len(points)
dp = [[0 for j in range(n)] for i in range(n)]
for i in range(n - 2, -1, -1):
for j in range(i + 2, n):
dp[i][j] = math.inf
for k in range(i + 1, j):
dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j] + triangle_area(points[i], points[j], points[k]))
return dp[0][n - 1]
# 测试
points = [(0, 0), (1, 0), (2, 1), (1, 2), (0, 2)]
print(optimal_triangulation(points))
```
伪代码:
```
function triangle_area(p1, p2, p3):
return abs(0.5 * ((p2[0] - p1[0]) * (p3[1] - p1[1]) - (p2[1] - p1[1]) * (p3[0] - p1[0])))
function optimal_triangulation(points):
n = length(points)
dp = array(n, n)
for i = n - 2 to 0 step -1:
for j = i + 2 to n - 1:
dp[i][j] = infinity
for k = i + 1 to j - 1:
dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j] + triangle_area(points[i], points[j], points[k]))
return dp[0][n - 1]
# 测试
points = [(0, 0), (1, 0), (2, 1), (1, 2), (0, 2)]
print(optimal_triangulation(points))
```
凸多边形最优三角剖分python代码注释
以下是凸多边形最优三角剖分的Python代码注释:
```
def optimal_triangulation(vertices):
"""
:param vertices: 凸多边形的顶点列表,按逆时针顺序排列
:return: 三角剖分的三角形列表,每个三角形由三个顶点的索引组成
"""
n = len(vertices)
dp = [[0] * n for _ in range(n)] # 初始化dp数组,dp[i][j]表示从第i个顶点到第j个顶点的最小三角剖分代价
# 枚举区间长度
for length in range(2, n):
# 枚举区间起点
for i in range(n - length):
# 区间终点
j = i + length
# 初始化dp[i][j]
dp[i][j] = float('inf')
# 枚举区间内的所有顶点,找到最优三角剖分
for k in range(i + 1, j):
dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j] + triangle_area(vertices[i], vertices[k], vertices[j]))
# 返回最优三角剖分
return construct_triangulation(0, n - 1, dp)
def triangle_area(p1, p2, p3):
"""
计算三角形面积
"""
return abs((p2[0] - p1[0]) * (p3[1] - p1[1]) - (p3[0] - p1[0]) * (p2[1] - p1[1])) / 2
def construct_triangulation(i, j, dp):
"""
构造最优三角剖分
"""
if j == i + 1:
return []
# 找到最优三角剖分的切分点
for k in range(i + 1, j):
if dp[i][j] == dp[i][k] + dp[k][j] + triangle_area(vertices[i], vertices[k], vertices[j]):
# 递归地构造左右两个子问题的最优三角剖分
left_triangulation = construct_triangulation(i, k, dp)
right_triangulation = construct_triangulation(k, j, dp)
# 将当前三角形加入最优三角剖分中
return left_triangulation + right_triangulation + [(i, k, j)]
```
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