多边形作为Vector2I对象列表(2维,整数坐标)给出。如何测试给定点是否在内?我在网上找到的所有实现都失败了一些微不足道的反例。写一个正确的实现似乎很难。语言并不重要,因为我会自己移植它。
答案 0 :(得分:23)
如果它是凸的,检查它的一个简单方法是该点位于所有段的同一侧(如果以相同的顺序遍历)。
你可以用十字产品轻松检查(因为它与段和点之间形成的角度的余弦成正比,带有正号的那些将位于右侧,而那些带有负号的位于左侧)。
以下是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 = x_product(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 x_product(a, b):
return a[0]*b[1]-a[1]*b[0]
答案 1 :(得分:14)
Ray Casting或Winding方法是此问题最常见的方法。有关详细信息,请参阅Wikipedia article。
另外,请查看this page以获取C语言中详细记录的解决方案。
答案 2 :(得分:7)
如果多边形是凸的,那么在C#中,以下实现&#34; test if always on same side&#34;方法,并且最多在O(多边形点的n)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 < polygon.Count - 1 ? i + 1 : 0;
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;
}
答案 3 :(得分:3)
openCV中的pointPolygonTest函数“确定该点是在轮廓内,外部还是位于边缘”: http://docs.opencv.org/modules/imgproc/doc/structural_analysis_and_shape_descriptors.html?highlight=pointpolygontest#pointpolygontest
答案 4 :(得分:3)
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
}
答案 5 :(得分:2)
我知道的方式是这样的。
你在多边形之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。
然后对于此多边形中的每一行,检查它们是否相交。
它们相交的线数总和给你内部与否。
如果是奇数:里面
如果是偶数:外面
答案 6 :(得分:1)
或者写这本书的人看到 - geometry page
特别是this page,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谁写了优秀的书Computational Geometry in C,它是Paul Bourke,但仍然是几何算法的非常好的来源。
答案 7 :(得分:1)
您必须检查要测试的点是否保持相对于凸多边形所有线段的方向。如果是这样,它就在里面。为此,对于每个段,请检查段矢量的行列式是否表示AB,而点矢量的行列式是否表示AP保留其符号。如果行列式为零,则该点在线段上。
要使用C#代码公开此内容,
save.execute
行列式演算,
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;
}
答案 8 :(得分:0)
这是我在项目中使用的版本。非常优雅简洁。适用于各种多边形。
http://www.eecs.umich.edu/courses/eecs380/HANDOUTS/PROJ2/InsidePoly.html
以下代码是兰道夫·富兰克林(Randolph Franklin)编写的,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;
}