python Graham求凸包问题并画图操作

python Graham求凸包并画图

python写Graham没有c++那么好写,但是python画图简单。只需要用matplotlib里的pyplot,c++画图太难了。

Graham算法写起来比较简单,只需要想办法对最小点和其他的点所连成的直线,与x轴正半轴的夹角进行排序,然后其他的就直接套用Graham算法模板就好了,因为c++可以重载排序函数sort,不用计算角度(用其他的数学方法),但是python不行(也许是我不知道而已,菜)。

python必须要在结构体里面加上角度这个变量,然后才能按照角度排序。排好序后就变得容易了,用stack栈存放答案,算完答案后,用scatter(散点图)画出点,用plt(折线图)画边界就好了。

import matplotlib.pyplot as plt
import math
import numpy as np  
class Node:
    def __init__(self):
        self.x = 0
        self.y = 0
        self.angel = 0
        #和最左下的点连成的直线,与x轴正半轴的夹角大小 
 
#按照角度从小到大排序
def cmp(x):
    return x.angel  
def bottom_point(points):
    min_index = 0
    n = len(points)
    #先判断y坐标,找出y坐标最小的点,x坐标最小的点
    for i in range(1, n):
        if points[i].y < points[min_index].y or (points[i].y == points[min_index].y and
           points[i].x < points[min_index].x):
            min_index = i
    return min_index 
 
#计算角度
def calc_angel(vec):
    norm = math.sqrt(vec[0] * vec[0] + vec[1] * vec[1])
    if norm == 0:
        return 0
    angel = math.acos(vec[0]/norm)
    if vec[1] >= 0:
        return angel
    else:
        return math.pi * 2 - angel 
 
def multi(v1, v2):
    return v1[0] * v2[1] - v1[1] * v2[0] 
 
point = []
n = 30
#生成30个点的坐标,n可以修改
for i in range(n):
    temp = Node()
    temp.x = np.random.randint(1, 100)
    temp.y = np.random.randint(1, 100)
    point.append(temp)
index = bottom_point(point)
for i in range(n):
    if i == index:
        continue
    #计算每个点和point[index]所连成的直线与x轴正半轴的夹角
    vector = [point[i].x - point[index].x, point[i].y - point[index].y]
    #vector是向量
    point[i].angel = calc_angel(vector)
#排序
point.sort(key=cmp)
#答案存入栈中
stack = []
stack.append(point[0])
stack.append(point[1])
#for循环更新答案
for i in range(2, n):
    L = len(stack)
    top = stack[L - 1]
    next_top = stack[L - 2]
    vec1 = [point[i].x - next_top.x, point[i].y - next_top.y]
    vec2 = [top.x - next_top.x, top.y - next_top.y]
    #一定要大于等于零,因为可能在一条直线上
    while multi(vec1, vec2) >= 0:
        stack.pop()
        L = len(stack)
        top = stack[L - 1]
        next_top = stack[L - 2]
        vec1 = [point[i].x - next_top.x, point[i].y - next_top.y]
        vec2 = [top.x - next_top.x, top.y - next_top.y]
    stack.append(point[i])
#画出图像
for p in point:
    plt.scatter(p.x, p.y, marker='o', c='g')
L = len(stack)
for i in range(L-1):
    plt.plot([stack[i].x, stack[i+1].x], [stack[i].y, stack[i+1].y], c='r')
plt.plot([stack[0].x, stack[L-1].x], [stack[0].y, stack[L-1].y], c='r')
plt.show()

Python 找到凸包 Convex hulls

图形学可以说经常遇到这东西了,这里给出一个库函数的实现

from scipy.spatial import ConvexHull
points = np.random.rand(10, 2) # 30 random points in 2-D
hull = ConvexHull(points)
import matplotlib.pyplot as plt
plt.plot(points[:,0], points[:,1], 'o')
for simplex in hull.simplices:
 plt.plot(points[simplex,0], points[simplex,1], 'k-')
plt.show()

以上为个人经验,希望能给大家一个参考,也希望大家多多支持脚本之家。

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