我在 iOS 上工作,有一个带有地理坐标的多边形,例如 (-27.589846, 151.982112)(-27.590174, 151.983045)(-27.590773, 151.982680)(-27.590602, 151.981908)。
我想知道它的外圆内圆:圆心和半径?
有什么办法吗?
谢谢?
最佳答案
您可以使用它来确定非自相交多边形的中心:
#include <iostream>
struct Point2D
{
double x;
double y;
};
Point2D compute2DPolygonCentroid(const Point2D* vertices, int vertexCount)
{
Point2D centroid = {0, 0};
double signedArea = 0.0;
double x0 = 0.0; // Current vertex X
double y0 = 0.0; // Current vertex Y
double x1 = 0.0; // Next vertex X
double y1 = 0.0; // Next vertex Y
double a = 0.0; // Partial signed area
// For all vertices except last
int i=0;
for (i=0; i<vertexCount-1; ++i)
{
x0 = vertices[i].x;
y0 = vertices[i].y;
x1 = vertices[i+1].x;
y1 = vertices[i+1].y;
a = x0*y1 - x1*y0;
signedArea += a;
centroid.x += (x0 + x1)*a;
centroid.y += (y0 + y1)*a;
}
// Do last vertex
x0 = vertices[i].x;
y0 = vertices[i].y;
x1 = vertices[0].x;
y1 = vertices[0].y;
a = x0*y1 - x1*y0;
signedArea += a;
centroid.x += (x0 + x1)*a;
centroid.y += (y0 + y1)*a;
signedArea *= 0.5;
centroid.x /= (6.0*signedArea);
centroid.y /= (6.0*signedArea);
return centroid;
}
int main()
{
Point2D polygon[] = {{0.0,0.0}, {0.0,10.0}, {10.0,10.0}, {10.0,0.0}};
size_t vertexCount = sizeof(polygon) / sizeof(polygon[0]);
Point2D centroid = compute2DPolygonCentroid(polygon, vertexCount);
std::cout << "Centroid is (" << centroid.x << ", " << centroid.y << ")\n";
}
要获得半径然后确定每个顶点的中心之间的距离并选择最大的一个!
关于ios - 计算多边形的最小内切圆,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/24140754/