多边形以 Vector2I 对象（二维，整数坐标）列表的形式给出。我如何测试给定点是否在内部？我在网上找到的所有实现都因一些微不足道的反例而失败。编写正确的实现似乎真的很难。语言无关紧要，因为我会自己移植。
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9 回答
31
如果它是凸的，检查它的简单方法是该点位于所有线段的同一侧（如果以相同的顺序遍历）。
您可以使用点积轻松检查（因为它与线段和点之间形成的角度的余弦成正比，如果我们用边缘的法线计算它，那些带有正号的将位于右侧并且左边有负号的）。
这是Python中的代码：
RIGHT = "RIGHT"
LEFT = "LEFT"
def inside_convex_polygon(point, vertices):
previous_side = None
n_vertices = len(vertices)
for n in xrange(n_vertices):
a, b = vertices[n], vertices[(n+1)%n_vertices]
affine_segment = v_sub(b, a)
affine_point = v_sub(point, a)
current_side = get_side(affine_segment, affine_point)
if current_side is None:
return False #outside or over an edge
elif previous_side is None: #first segment
previous_side = current_side
elif previous_side != current_side:
return False
return True
def get_side(a, b):
x = cosine_sign(a, b)
if x < 0:
return LEFT
elif x > 0:
return RIGHT
else:
return None
def v_sub(a, b):
return (a[0]-b[0], a[1]-b[1])
def cosine_sign(a, b):
return a[0]*b[1]-a[1]*b[0]
于 2009-07-13T14:11:57.473 回答
13
如果多边形是凸的，那么在 C# 中，以下实现了“测试是否始终在同一侧”方法，并且最多在 O(n 个多边形点)处运行：
public static bool IsInConvexPolygon(Point testPoint, List<Point> polygon)
{
//Check if a triangle or higher n-gon
Debug.Assert(polygon.Length >= 3);
//n>2 Keep track of cross product sign changes
var pos = 0;
var neg = 0;
for (var i = 0; i < polygon.Count; i++)
{
//If point is in the polygon
if (polygon[i] == testPoint)
return true;
//Form a segment between the i'th point
var x1 = polygon[i].X;
var y1 = polygon[i].Y;
//And the i+1'th, or if i is the last, with the first point
var i2 = (i+1)%polygon.Count;
var x2 = polygon[i2].X;
var y2 = polygon[i2].Y;
var x = testPoint.X;
var y = testPoint.Y;
//Compute the cross product
var d = (x - x1)*(y2 - y1) - (y - y1)*(x2 - x1);
if (d > 0) pos++;
if (d < 0) neg++;
//If the sign changes, then point is outside
if (pos > 0 && neg > 0)
return false;
}
//If no change in direction, then on same side of all segments, and thus inside
return true;
}
于 2016-01-09T03:01:57.533 回答
3
openCV 中的 pointPolygonTest 函数“确定该点是在轮廓内部、外部还是位于边缘”： http ://docs.opencv.org/modules/imgproc/doc/structural_analysis_and_shape_descriptors.html?highlight=pointpolygontest#pointpolygontest
于 2013-01-02T01:06:28.260 回答
3
fortran 的答案几乎对我有用，除了我发现我必须翻译多边形以便您测试的点与原点相同。这是我编写的用于完成这项工作的 JavaScript：
function Vec2(x, y) {
return [x, y]
}
Vec2.nsub = function (v1, v2) {
return Vec2(v1[0]-v2[0], v1[1]-v2[1])
}
// aka the "scalar cross product"
Vec2.perpdot = function (v1, v2) {
return v1[0]*v2[1] - v1[1]*v2[0]
}
// Determine if a point is inside a polygon.
//
// point - A Vec2 (2-element Array).
// polyVerts - Array of Vec2's (2-element Arrays). The vertices that make
// up the polygon, in clockwise order around the polygon.
//
function coordsAreInside(point, polyVerts) {
var i, len, v1, v2, edge, x
// First translate the polygon so that `point` is the origin. Then, for each
// edge, get the angle between two vectors: 1) the edge vector and 2) the
// vector of the first vertex of the edge. If all of the angles are the same
// sign (which is negative since they will be counter-clockwise) then the
// point is inside the polygon; otherwise, the point is outside.
for (i = 0, len = polyVerts.length; i < len; i++) {
v1 = Vec2.nsub(polyVerts[i], point)
v2 = Vec2.nsub(polyVerts[i+1 > len-1 ? 0 : i+1], point)
edge = Vec2.nsub(v1, v2)
// Note that we could also do this by using the normal + dot product
x = Vec2.perpdot(edge, v1)
// If the point lies directly on an edge then count it as in the polygon
if (x < 0) { return false }
}
return true
}
于 2013-03-09T08:02:03.527 回答
2
我知道的方式是这样的。
您在多边形外的某处选择一个点，它可能远离几何体。然后你从这一点画一条线。我的意思是你用这两点创建一个线方程。
然后对于这个多边形中的每条线，检查它们是否相交。
它们相交线数的总和给你它是否在里面。
如果它是奇怪的：里面
如果是偶数：外面
于 2009-07-13T14:38:45.637 回答
2
您必须检查要测试的点是否保持其相对于凸多边形所有段的方向。如果是这样，它在里面。要对每个段执行此操作，请检查段向量的行列式是否为 AB 和点的向量是否为 AP 保留它的符号。如果行列式为零，则该点位于该段上。
要在 C# 代码中公开这一点，
public bool IsPointInConvexPolygon(...)
{
Point pointToTest = new Point(...);
Point pointA = new Point(...);
....
var polygon = new List<Point> { pointA, pointB, pointC, pointD ... };
double prevPosition = 0;
// assuming polygon is convex.
for (var i = 0; i < polygon.Count; i++)
{
var startPointSegment = polygon[i];
// end point is first point if the start point is the last point in the list
// (closing the polygon)
var endPointSegment = polygon[i < polygon.Count - 1 ? i + 1 : 0];
if (pointToTest.HasEqualCoordValues(startPointSegment) ||
pointToTest.HasEqualCoordValues(endPointSegment))
return true;
var position = GetPositionRelativeToSegment(pointToTest, startPointSegment, endPointSegment);
if (position == 0) // point position is zero so we are on the segment, we're on the polygon.
return true;
// after we checked the test point's position relative to the first segment, the position of the point
// relative to all other segments must be the same as the first position. If not it means the point
// is not inside the convex polygon.
if (i > 0 && prevPosition != position)
return false;
prevPosition = position;
}
return true;
}
行列式计算，
public double GetPositionRelativeToSegment(Point pointToTest, Point segmentStart, Point segmentEnd)
{
return Math.Sign((pointToTest.X - segmentStart.X) * (segmentEnd.Y - segmentStart.Y) -
(pointToTest.Y - segmentStart.Y) * (segmentEnd.X - segmentStart.X));
}
于 2020-03-25T07:25:45.383 回答
1
或者从写这本书的人那里看到 -几何页面
特别是这个页面，他讨论了为什么缠绕规则通常比射线交叉更好。
编辑 - 抱歉，这不是Jospeh O'Rourke写的优秀著作Computational Geometry in C，它是 Paul Bourke，但仍然是几何算法的非常好的来源。
于 2009-07-13T14:13:36.003 回答
0
这是我在项目中使用的版本。它非常优雅和简洁。适用于各种多边形。
http://www.eecs.umich.edu/courses/eecs380/HANDOUTS/PROJ2/InsidePoly.html
以下代码由 Randolph Franklin 编写，它为内部点返回 1，为外部点返回 0。
int pnpoly(int npol, float *xp, float *yp, float x, float y)
{
int i, j, c = 0;
for (i = 0, j = npol-1; i < npol; j = i++) {
if ((((yp[i] <= y) && (y < yp[j])) ||
((yp[j] <= y) && (y < yp[i]))) &&
(x < (xp[j] - xp[i]) * (y - yp[i]) / (yp[j] - yp[i]) + xp[i]))
c = !c;
}
return c;
}
于 2020-12-11T07:24:55.600 回答