python寫Graham沒有c++那么好寫，但是python畫圖簡單。只需要用matplotlib里的pyplot，c++畫圖太難了。

Graham算法寫起來比較簡單，只需要想辦法對最小點(diǎn)和其他的點(diǎn)所連成的直線，與x軸正半軸的夾角進(jìn)行排序，然后其他的就直接套用Graham算法模板就好了，因?yàn)閏++可以重載排序函數(shù)sort，不用計(jì)算角度（用其他的數(shù)學(xué)方法），但是python不行（也許是我不知道而已，菜）。

python必須要在結(jié)構(gòu)體里面加上角度這個(gè)變量，然后才能按照角度排序。排好序后就變得容易了，用stack棧存放答案，算完答案后，用scatter（散點(diǎn)圖）畫出點(diǎn)，用plt（折線圖）畫邊界就好了。

import matplotlib.pyplot as pltimport mathimport numpy as np class Node: def __init__(self):self.x = 0self.y = 0self.angel = 0#和最左下的點(diǎn)連成的直線，與x軸正半軸的夾角大小 #按照角度從小到大排序def cmp(x): return x.angel def bottom_point(points): min_index = 0 n = len(points) #先判斷y坐標(biāo)，找出y坐標(biāo)最小的點(diǎn)，x坐標(biāo)最小的點(diǎn) 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 #計(jì)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個(gè)點(diǎn)的坐標(biāo)，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 #計(jì)算每個(gè)點(diǎn)和point[index]所連成的直線與x軸正半軸的夾角 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)#答案存入棧中stack = []stack.append(point[0])stack.append(point[1])#for循環(huán)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?yàn)榭赡茉谝粭l直線上 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])#畫出圖像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

圖形學(xué)可以說經(jīng)常遇到這東西了，這里給出一個(gè)庫函數(shù)的實(shí)現(xiàn)

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

以上為個(gè)人經(jīng)驗(yàn)，希望能給大家一個(gè)參考，也希望大家多多支持好吧啦網(wǎng)。