求任意多边形面积 python实现

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数学解决方法:

多边形外选取一点，连接各点构成三角形，计算求和...... 详细链接 http://blog.csdn.net/hemmingway/article/details/7814494

已知三角形三边长，求三角形面积------>海伦公式链接:http://www.zybang.com/question/29d209e9732d1f1f6a6de35b94edd3ba.html

由上实现下面python代码源代码出处链接: http://blog.csdn.net/zfq43/article/details/8595288

# -*- coding: cp936 -*-
import mathclass Point():
def __init__(self,x,y):
self.x = x
self.y = ydef GetAreaOfPolyGon(points): area = 0
if(len(points)<3): raise Exception("error") p1 = points[0]
for i in range(1,len(points)-1):
p2 = points[1]
p3 = points[2] #计算向量
vecp1p2 = Point(p2.x-p1.x,p2.y-p1.y)
vecp2p3 = Point(p3.x-p2.x,p3.y-p2.y) #判断顺时针还是逆时针，顺时针面积为正，逆时针面积为负
vecMult = vecp1p2.x*vecp2p3.y - vecp1p2.y*vecp2p3.x #判断正负方向比较有意思
sign = 0
if(vecMult>0):
sign = 1
elif(vecMult<0):
sign = -1 triArea = GetAreaOfTriangle(p1,p2,p3)*sign
area += triArea
return abs(area)def GetAreaOfTriangle(p1,p2,p3):
'''计算三角形面积海伦公式'''
area = 0
p1p2 = GetLineLength(p1,p2)
p2p3 = GetLineLength(p2,p3)
p3p1 = GetLineLength(p3,p1)
s = (p1p2 + p2p3 + p3p1)/2
area = s*(s-p1p2)*(s-p2p3)*(s-p3p1) #海伦公式
area = math.sqrt(area)
return areadef GetLineLength(p1,p2):
'''计算边长'''
length = math.pow((p1.x-p2.x),2) + math.pow((p1.y-p2.y),2) #pow 次方
length = math.sqrt(length)
return length def main():
p1 = Point(1,1)
p2 = Point(2,1)
p3 = Point(2,2)
p4 = Point(1,2)
points = [p1,p2,p3,p4]
area = GetAreaOfPolyGon(points)
print(math.ceil(area))
assert math.ceil(area)==1if __name__ == '__main__':
main()