Like David Segonds' answer suggests I use an approach of angle summation derived from my concave polygon drawing algorithm. It relies of adding up the approximate angles of subtriangles around the point to obtain a weight. A weight around 1.0 means the point is inside the triangle, a weight around 0.0 means outside, a weight around -1.0 is what happens when inside the polygon but in reverse order (like with one of the halves of a bowtie-shaped tetragon) and a weight of NAN if exactly on an edge. The reason it's not slow is that angles don't need to be estimated accurately at all. Holes can be handled by treating them as separate polygons and subtracting the weights.

typedef struct { double x, y; } xy_t;

xy_t sub_xy(xy_t a, xy_t b)
{
    a.x -= b.x;
    a.y -= b.y;
    return a;
}

double calc_sharp_subtriangle_pixel_weight(xy_t p0, xy_t p1)
{
    xy_t rot, r0, r1;
    double weight;

    // Rotate points (unnormalised)
    rot = sub_xy(p1, p0);
    r0.x = rot.x*p0.y - rot.y*p0.x;
    r0.y = rot.x*p0.x + rot.y*p0.y;
    r1.y = rot.x*p1.x + rot.y*p1.y;

    // Calc weight
    weight = subtriangle_angle_approx(r1.y, r0.x) - subtriangle_angle_approx(r0.y, r0.x);

    return weight;
}

double calc_sharp_polygon_pixel_weight(xy_t p, xy_t *corner, int corner_count)
{
    int i;
    xy_t p0, p1;
    double weight = 0.;

    p0 = sub_xy(corner[corner_count-1], p);
    for (i=0; i < corner_count; i++)
    {
        // Transform corner coordinates
        p1 = sub_xy(corner[i], p);

        // Calculate weight for each subtriangle
        weight += calc_sharp_subtriangle_pixel_weight(p0, p1);
        p0 = p1;
    }

    return weight;
}

因此，对于多边形的每一段，都形成一个子三角形，并计算点，然后旋转每个子三角形以计算其近似角度并添加到权重。

调用subtriangle_angle_approx(y, x)可以替换为atan2(y, x) / (2.*pi)，但是一个非常粗略的近似值就足够精确了:

double subtriangle_angle_approx(double y, double x)
{
    double angle, d;
    int obtuse;

    if (x == 0.)
        return NAN;

    obtuse = fabs(y) > fabs(x);
    if (obtuse)
        swap_double(&y, &x);

    // Core of the approximation, a very loosely approximate atan(y/x) / (2.*pi) over ]-1 , 1[
    d = y / x;
    angle = 0.13185 * d;

    if (obtuse)
        angle = sign(d)*0.25 - angle;

    return angle;
}

在Ray casting算法中处理以下特殊情况:

射线与多边形的一条边重叠。 点在多边形的内部，光线穿过多边形的顶点。 该点在多边形的外部，光线只接触到多边形的一个角。

检查确定一个点是否在一个复杂多边形内。本文提供了一种简单的解决方法，因此对于上述情况不需要特殊处理。

没有什么比归纳定义问题更美好的了。为了完整起见，你在序言中有一个版本，它可能也澄清了光线投射背后的思想:

基于仿真的简化算法在http://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html

一些helper谓词:

exor(A,B):- \+A,B;A,\+B.
in_range(Coordinate,CA,CB) :- exor((CA>Coordinate),(CB>Coordinate)).

inside(false).
inside(_,[_|[]]).
inside(X:Y, [X1:Y1,X2:Y2|R]) :- in_range(Y,Y1,Y2), X > ( ((X2-X1)*(Y-Y1))/(Y2-Y1) +      X1),toggle_ray, inside(X:Y, [X2:Y2|R]); inside(X:Y, [X2:Y2|R]).

get_line(_,_,[]).
get_line([XA:YA,XB:YB],[X1:Y1,X2:Y2|R]):- [XA:YA,XB:YB]=[X1:Y1,X2:Y2]; get_line([XA:YA,XB:YB],[X2:Y2|R]).

给定两点a和B的直线(直线(a,B))方程为:

                    (YB-YA)
           Y - YA = ------- * (X - XA) 
                    (XB-YB) 

It is important that the direction of rotation for the line is setted to clock-wise for boundaries and anti-clock-wise for holes. We are going to check whether the point (X,Y), i.e the tested point is at the left half-plane of our line (it is a matter of taste, it could also be the right side, but also the direction of boundaries lines has to be changed in that case), this is to project the ray from the point to the right (or left) and acknowledge the intersection with the line. We have chosen to project the ray in the horizontal direction (again it is a matter of taste, it could also be done in vertical with similar restrictions), so we have:

               (XB-XA)
           X < ------- * (Y - YA) + XA
               (YB-YA) 

Now we need to know if the point is at the left (or right) side of the line segment only, not the entire plane, so we need to restrict the search only to this segment, but this is easy since to be inside the segment only one point in the line can be higher than Y in the vertical axis. As this is a stronger restriction it needs to be the first to check, so we take first only those lines meeting this requirement and then check its possition. By the Jordan Curve theorem any ray projected to a polygon must intersect at an even number of lines. So we are done, we will throw the ray to the right and then everytime it intersects a line, toggle its state. However in our implementation we are goint to check the lenght of the bag of solutions meeting the given restrictions and decide the innership upon it. for each line in the polygon this have to be done.

is_left_half_plane(_,[],[],_).
is_left_half_plane(X:Y,[XA:YA,XB:YB], [[X1:Y1,X2:Y2]|R], Test) :- [XA:YA, XB:YB] =  [X1:Y1, X2:Y2], call(Test, X , (((XB - XA) * (Y - YA)) / (YB - YA) + XA)); 
                                                        is_left_half_plane(X:Y, [XA:YA, XB:YB], R, Test).

in_y_range_at_poly(Y,[XA:YA,XB:YB],Polygon) :- get_line([XA:YA,XB:YB],Polygon),  in_range(Y,YA,YB).
all_in_range(Coordinate,Polygon,Lines) :- aggregate(bag(Line),    in_y_range_at_poly(Coordinate,Line,Polygon), Lines).

traverses_ray(X:Y, Lines, Count) :- aggregate(bag(Line), is_left_half_plane(X:Y, Line, Lines, <), IntersectingLines), length(IntersectingLines, Count).

% This is the entry point predicate
inside_poly(X:Y,Polygon,Answer) :- all_in_range(Y,Polygon,Lines), traverses_ray(X:Y, Lines, Count), (1 is mod(Count,2)->Answer=inside;Answer=outside).

计算点p与每个多边形顶点之间的有向角和。如果总倾斜角是360度，那么这个点在里面。如果总数为0，则点在外面。

我更喜欢这种方法，因为它更健壮，对数值精度的依赖更小。

计算交集数量的均匀性的方法是有限的，因为你可以在计算交集数量的过程中“击中”一个顶点。

编辑:顺便说一下，这种方法适用于凹凸多边形。

编辑:我最近在维基百科上找到了一篇关于这个话题的完整文章。

对于检测多边形上的命中，我们需要测试两件事:

如果点在多边形区域内。(可通过Ray-Casting算法实现) 如果点在多边形边界上(可以用与在折线(线)上检测点相同的算法来完成)。

在C语言的多边形测试中，有一个点没有使用光线投射。它可以用于重叠区域(自我交叉)，请参阅use_holes参数。

/* math lib (defined below) */
static float dot_v2v2(const float a[2], const float b[2]);
static float angle_signed_v2v2(const float v1[2], const float v2[2]);
static void copy_v2_v2(float r[2], const float a[2]);

/* intersection function */
bool isect_point_poly_v2(const float pt[2], const float verts[][2], const unsigned int nr,
                         const bool use_holes)
{
    /* we do the angle rule, define that all added angles should be about zero or (2 * PI) */
    float angletot = 0.0;
    float fp1[2], fp2[2];
    unsigned int i;
    const float *p1, *p2;

    p1 = verts[nr - 1];

    /* first vector */
    fp1[0] = p1[0] - pt[0];
    fp1[1] = p1[1] - pt[1];

    for (i = 0; i < nr; i++) {
        p2 = verts[i];

        /* second vector */
        fp2[0] = p2[0] - pt[0];
        fp2[1] = p2[1] - pt[1];

        /* dot and angle and cross */
        angletot += angle_signed_v2v2(fp1, fp2);

        /* circulate */
        copy_v2_v2(fp1, fp2);
        p1 = p2;
    }

    angletot = fabsf(angletot);
    if (use_holes) {
        const float nested = floorf((angletot / (float)(M_PI * 2.0)) + 0.00001f);
        angletot -= nested * (float)(M_PI * 2.0);
        return (angletot > 4.0f) != ((int)nested % 2);
    }
    else {
        return (angletot > 4.0f);
    }
}

/* math lib */

static float dot_v2v2(const float a[2], const float b[2])
{
    return a[0] * b[0] + a[1] * b[1];
}

static float angle_signed_v2v2(const float v1[2], const float v2[2])
{
    const float perp_dot = (v1[1] * v2[0]) - (v1[0] * v2[1]);
    return atan2f(perp_dot, dot_v2v2(v1, v2));
}

static void copy_v2_v2(float r[2], const float a[2])
{
    r[0] = a[0];
    r[1] = a[1];
}

注意:这是一个不太理想的方法，因为它包含很多对atan2f的调用，但它可能会引起阅读这个线程的开发人员的兴趣(在我的测试中，它比使用线交方法慢23倍)。