圆形路径算法

Question

我一直在开发一种循环路径算法，以创建一条出点的路径。 这是我开始的数组：

(1,1)
(1,6)
(2,2)
(2,5)
(4,1)
(4,2)
(6,5)
(6,6)

这些是坐标系中的点，我想要排序，所以我只需要相邻点之间的水平线或垂直线。所以排序后的数组需要是这样的

(1,1)        (A,H)
(1,6)        (A,B)
(6,6)        (C,B)
(6,5)        (C,D)
(2,5)        (E,D)
(2,2)        (E,F)
(4,2)        (G,F)
(4,1)        (G,H)

编辑：这些点是从不同的边缘中提取的。每条边由两个点定义。没有相互重叠的边缘。

Edge: (1,1) -> (1,6)
Edge: (1,6) -> (6,6)
Edge: (6,6) -> (6,5)
Edge: (6,5) -> (2,5)
Edge: (2,5) -> (2,2)
Edge: (2,2) -> (4,2)
Edge: (4,2) -> (4,1)
Edge: (4,1) -> (1,1)

感谢您的帮助！

Answer 1

假设，边缘必须交替（每个垂直后面跟着水平，反之亦然），算法可能如下：

P = input // collection of points
EH = []   // collection of horizontal edges
EV = []   // collection of vertical edges

sort P by x, then y         // (1,1), (1,2), (1,4), (1,6), (2,2), (2,5), ...
for (i = 0; i < P.size; i += 2) 
    if P[i].x != P[i+1].x return   // no solution
    EH.add(new edge(P[i], P[i+1]))

sort P by y, then x         // (1,1), (4,1), (2,2), (4,2), ...
for (i = 0; i < P.size; i += 2)
    if P[i].y != P[i+1].y return   // no solution
    EV.add(new edge(P[i], P[i+1]))

// After this, every point belongs to 1 horizontal egde and 1 vertical
// If exists closed path which traverses all points without overlapping, 
// such path is formed by these edges

S = []          // path
S.add(EH[0])
cp = EH[0].p2   // closing point 
p =  EH[0].p1   // current ending point
find edge e in EV where e.p1 = p or e.p2 = p
if e not found return empty path      // no solution
S.add(e)   
if p = e.p1
    p = e.p2
else
    p = e.p1
while (p != cp) {
    find edge e in EH where e.p1 = p or e.p2 = p
    if not found return empty path    // no solution
    S.add(e)
    if p = e.p1           
       p = e.p2
    else
       p = e.p1
    find edge e in EV where e.p1 = p or e.p2 = p
    if not found return empty path    // no solution
    S.add(e)
    if p = e.p1 
       p = e.p2
    else
       p = e.p1
}
return S 

为了简化边搜索，提到的集合 EV 和 EH 可以使用哈希映射（点、点）实现，其中对于每个实际边 e(p1, p2) 应放置两个条目：p1->p2 和 p2->p1 .每个点都属于 1 个水平和 1 个垂直边缘，因此条目不会被覆盖。因此可以简化算法的第二部分：

while (p != cp) {
    get from EH point pp by key p
    if not found return empty path    
    S.add(new edge(p, pp))
    p = pp
    get from EV point pp by key p
    if not found return empty path    
    S.add(new edge(p, pp))
    p = pp
}