Question

我有一对（闭合的）多边形，每个多边形定义为一系列点（顶点）。多边形各自代表一块土地，由一条小河分隔，因此溪流在两个多边形之间形成一个狭窄的间隙。

我正在寻找一种算法，通过将两个多边形连接成一个连接的多边形来识别和消除间隙。

下图显示了一个示例，其中原始多边形为绿色和红色，生成的多边形以黄色显示。

example

到目前为止，我已经能够做到以下几点：

对于polygon-A中的每条边，在polygon-B中找到最近的顶点。
查找polygon-B的所有顶点，这些顶点位于polygon-A的特定距离内。

但我不确定我现在需要做什么。

Answer 1

为了完整起见，我想分享我的解决方案，作为python实现。这是基于Retsam接受的答案，以及David Eisenstat在对该答案的评论中提出的想法，即替换连接到同一原始多边形上的顶点的凸包上的边缘，以及来自该多边形的插入顶点

def joinPolygons(polya, polyb):
    """
    Generate and return a single connected polygon which includes the two given
    polygons. The connection between the two polygons is based on the convex hull
    of the composite polygon. All polygons are sequences of two-tuples giving the
    vertices of the polygon as (x, y), in order. That means vertices that are adjacent
    in the sequence are adjacent in the polygon (connected by an edge). The first and
    last vertices in the sequence are also connected by any edge (implicitly closed, do
    not duplicate the first point at the end of the sequence to close it).

    Only simple polygons are supported (no self-intersection).
    """

    #Just to make it easier to identify and access by index.
    polygons = [polya, polyb]

    #Create a single list of points to create the convex hull for (each
    # point is a vertex of one of the polygons).
    #Additionally, each point includes some additional "cargo", indicating which
    # polygon it's from, and it's index into that polygon
    # This assumes the implementation of convexHull simply ignores everything
    # beyond the first two elements of each vertex.
    composite = []
    for i in range(len(polygons)):
        points = polygons[i]
        composite += [(points[j][0], points[j][1], j, i) for j in xrange(len(points))]

    #Get the convex hull of the two polygons together.
    ch = convexHull(composite)

    #Now we're going to walk along the convex hull and find edges that connect two vertices
    # from the same source polygon. We then replace that edge with all the intervening edges
    # from that source polygon.

    #Start with the first vertex in the CH.
    x, y, last_vnum, last_pnum = ch[0]

    #Here is where we will collect the vertices for our resulting polygon, starting with the
    # first vertex on the CH (all the vertices on the CH will end up in the result, plus some
    # additional vertices from the original polygons).
    results = [(x, y)]

    #The vertices of the convex hull will always walk in a particular direction around each
    # polygon (i.e., forwards in the sequence of vertices, or backwards). We will use this
    # to keep track of which way they go.
    directions = [None for poly in polygons]

    #Iterate over all the remaining points in the CH, and then back to the first point to
    # close it.
    for x, y, vnum, pnum in list(ch[1:]) + [ch[0]]:

        #If this vertex came from the same original polygon as the last one, we need to
        # replace the edge between them with all the intervening edges from that polygon.
        if pnum == last_pnum:

            #Number of vertices in the polygon
            vcount = len(polygons[pnum])

            #If an edge of the convex hull connects the first and last vertex of the polygon,
            # then the CH edge must also be an edge of the polygon, because the two vertices are
            # adjacent in the original polygon. Therefore, if the convex
            # hull goes from the first vertex to the last in a single edge, then it's walking
            # backwards around the polygon. Likewise, if it goes from the last to the first in 
            # a single edge, it's walking forwards.
            if directions[pnum] is None:
                if last_vnum < vnum:
                    if last_vnum == 0 and vnum == vcount - 1:
                        direction = -1
                    else:
                        direction = 1
                else:
                    if last_vnum == vcount - 1 and vnum == 0:
                        direction = 1
                    else:
                        direction = -1
                directions[pnum] = direction
            else:
                direction = directions[pnum]

            #Now walk from the previous vertex to the current one on the source
            # polygon, and add all the intevening vertices (as well as the current one
            # from the CH) onto the result.
            v = last_vnum
            while v != vnum:
                v += direction
                if v >= vcount:
                    v = 0
                elif v == -1:
                    v = vcount - 1
                results.append(polygons[pnum][v])

        #This vertex on the CH is from a different polygon originally than the previous
        # vertex, so we just leave them connected.
        else:
            results.append((x, y))

        #Remember this vertex for next time.
        last_vnum = vnum
        last_pnum = pnum

    return results



def convexHull(points, leftMostVert=None):
    """
    Returns a new polygon which is the convex hull of the given polygon.

    :param: leftMostVert    The index into points of the left most vertex in the polygon.
                            If you don't know what it is, pass None and we will figure it
                            out ourselves.
    """
    point_count = len(points)

    #This is implemented using the simple Jarvis march "gift wrapping" algorithm.
    # Generically, to find the next point on the convex hull, we find the point
    # which has the smallest clockwise-angle from the previous edge, around the
    # last point. We start with the left-most point and a virtual vertical edge
    # leading to it.

    #If the left-most vertex wasn't specified, find it ourselves.
    if leftMostVert is None:
        minx = points[0][0]
        leftMostVert = 0
        for i in xrange(1, point_count):
            x = points[i][0]
            if x < minx:
                minx = x
                leftMostVert = i

    #This is where we will build up the vertices we want to include in the hull.
    # They are stored as indices into the sequence `points`.
    sel_verts = [leftMostVert]

    #This is information we need about the "last point" and "last edge" in order to find
    # the next point. We start with the left-most point and a pretend vertical edge.

    #The index into `points` of the last point.
    sidx = leftMostVert

    #The actual coordinates (x,y) of the last point.
    spt = points[sidx]

    #The vector of the previous edge.
    # Vectors are joined tail to tail to measure angle, so it
    # starts at the last point and points towards the previous point.
    last_vect = (0, -1, 0)
    last_mag = 1.0

    #Constant
    twopi = 2.0*math.pi

    #A range object to iterate over the vertex numbers.
    vert_nums = range(point_count)

    #A list of indices of points which have been determined to be colinear with
    # another point and a selected vertex on the CH, and which are not endpoints
    # of the line segment. These points are necessarily not vertices of the convex
    # hull: at best they are internal to one of its edges.
    colinear = []

    #Keep going till we come back around to the first (left-most) point.
    while True:
        #Try all other end points, find the one with the smallest CW angle.
        min_angle = None
        for i in vert_nums:

            #Skip the following points:
            # -The last vertex (sidx)
            # -The second to last vertex (sel_verts[-2]), that would just backtrack along
            #  the edge we just created.
            # -Any points which are determined to be colinear and internal (indices in `colinear`).
            if i == sidx or (len(sel_verts) > 1 and i == sel_verts[-2]) or i in colinear:
                continue

            #The point to test (x,y)
            pt = points[i]

            #vector from current point to test point.
            vect = (pt[0] - spt[0], pt[1] - spt[1], 0)
            mag = math.sqrt(vect[0]*vect[0] + vect[1]*vect[1])

            #Now find clockwise angle between the two vectors. Start by
            # finding the smallest angle between them, using the dot product.
            # Then use cross product and right-hand rule to determine if that
            # angle is clockwise or counter-clockwise, and adjust accordingly.

            #dot product of the two vectors.
            dp = last_vect[0]*vect[0] + last_vect[1]*vect[1]
            cos_theta = dp / (last_mag * mag)

            #Ensure fp erros don't become domain errors.
            if cos_theta > 1.0:
                cos_theta = 1.0
            elif cos_theta < -1.0:
                cos_theta = -1.0

            #Smaller of the two angles between them.
            theta = math.acos(cos_theta)

            #Take cross product of last vector by test vector.
            # Except we know that Z components in both input vectors are 0,
            # So the X and Y components of the resulting vector will be 0. Plus,
            # we only care aboue the Z component of the result.
            cpz = last_vect[0]*vect[1] - last_vect[1]*vect[0]

            #Assume initially that angle between the vectors is clock-wise.
            cwangle = theta
            #If the cross product points up out of the plane (positive Z),
            # then the angle is actually counter-clockwise.
            if cpz > 0:
                cwangle = twopi - theta

            #If this point has a smaller angle than the others we've considered,
            # choose it as the new candidate.
            if min_angle is None or cwangle < min_angle:
                min_angle = cwangle
                next_vert = i
                next_mvect = vect
                next_mag = mag
                next_pt = pt

            #If the angles are the same, then they are colinear with the last vertex. We want
            # to pick the one which is furthest from the vertex, and put all other colinear points
            # into the list so we can skip them in the future (this isn't just an optimization, it
            # appears to be necessary, otherwise we will pick one of the other colinear points as
            # the next vertex, which is incorrect).
            #Note: This is fine even if this turns out to be the next edge of the CH (i.e., we find
            # a point with a smaller angle): any point with is internal-colinear will not be a vertex
            # of the CH.
            elif cwangle == min_angle:
                if mag > next_mag:
                    #This one is further from the last vertex, so keep it as the candidate, and put the
                    # other colinear point in the list.
                    colinear.append(next_vert)
                    min_angle = cwangle
                    next_vert = i
                    next_mvect = vect
                    next_mag = mag
                    next_pt = pt
                else:
                    #This one is closer to the last vertex than the current candidate, so just keep that
                    # as the candidate, and put this in the list.
                    colinear.append(i)

        #We've found the next vertex on the CH.
        # If it's the first vertex again, then we're done.
        if next_vert == leftMostVert:
            break
        else:
            #Otherwise, add it to the list of vertices, and mark it as the
            # last vertex.
            sel_verts.append(next_vert)
            sidx = next_vert
            spt = next_pt
            last_vect = (-next_mvect[0], -next_mvect[1])
            last_mag = next_mag

    #Now we have a list of vertices into points, but we really want a list of points, so
    # create that and return it.
    return tuple(points[i] for i in sel_verts)

Answer 2

您可能会看一下凸包算法的修改。凸包算法采用一组点并绘制包含这些点的最小凸形。你的问题几乎是一个凸壳问题，除了顶部的那些凹陷区域。简单地使用凸包算法会给你这个，这很接近，但不是你需要的（注意。棕色区域是不同的）

Convex Hull

根据您要做的事情，凸包可能“足够好”，但如果没有，您仍然可以修改算法以忽略非凸的部分，并简化合并两个多边形。

具体而言this pdf显示了如何合并两个凸包，这与你想要做的很相似。

Answer 3

您可以尝试morphological operations。具体来说，您可以尝试扩张，然后尝试侵蚀（也称为形态“关闭”）。 n 像素的膨胀 - 其中 n 大于河流的宽度 - 将组合形状。随后的侵蚀将消除对该图其余部分造成的大部分损害。它不是完美的（将完美地结合两种形状，但代价是其他形状的某些软化），但也许是由于总操作的结果，你可以想象找出纠正它的方法。

通常，这些形态学操作是在位图上完成的，而不是在多边形上完成的。但是在多边形的角上运行简单的操作可能会有效。

Answer 4

这是一种粗略，可扩展的方法。

使用大小为Z-by-Z的单元格将您的空间量化为网格，并将每个单元格命名为其索引（i，j）。
对于每个多边形P，对于每个顶点V，用包含它的单元格（i，j）标识（P，V）。
对于每个单元格（i，j），考虑已经用它识别的多边形顶点对（P_k，V_k），k = 1 ... K的集合。当且仅当P_a和P_b不是相同的多边形并且V_a和V_b在这些对中最接近时，合并多边形P_a和P_b的顶点V_a和V_b。重复合并顶点，直到不再合并为止。合并的顶点获取源顶点位置的平均值。

Answer 5

使用Boost.Geometry，您可以以正距离buffer调用boost::geometry::strategy::buffer::distance_symmetric<double>{r}，正距离buffer足以弥合绿色和红色多边形之间的间隙，然后用{{1 }}。如果第二次调用产生伪像，请向boost::geometry::strategy::buffer::distance_symmetric<double>{-r}添加一个非常小的数字，例如r

结合附近的多边形

5 个答案: