概述
deltas = [[1,-1],[0,1,[-1,0],1],-1,[1,0]]
class HexGrid():
def __init__(self,radius):
self.radius = radius
self.tiles = {(0,0): "X"}
for r in range(radius):
a = 0
b = -r
c = +r
for j in range(6):
num_of_hexas_in_edge = r
for i in range(num_of_hexas_in_edge):
a = a+deltas[j][0]
b = b+deltas[j][1]
c = c+deltas[j][2]
self.tiles[a,b,c] = "X"
def show(self):
l = []
for y in range(20):
l.append([])
for x in range(60):
l[y].append(".")
for (a,c),tile in self.tiles.iteritems():
l[self.radius-1-b][a-c+(2*(self.radius-1))] = self.tiles[a,c]
mapString = ""
for y in range(len(l)):
for x in range(len(l[y])):
mapString += l[y][x]
mapString += "\n"
print(mapString)import hexgrid hg = hexgrid.HexGrid(radius)
并访问这样的坐标：
hg.tiles[a,c]
这似乎现在工作正常,但我确信以这种方式存储地图肯定存在一些缺点.如果有任何缺点,请指出它们,并提出更好的存储方式？非常感谢你的时间.
但是,您错过了一种管理此类地图的简单方法.将其视为行和列,只是单元格的形状略有不同.
+--+--+--+--+--+--+--+ \/ \/ \/ \/ \/ \/ \/ Even row /\ /\ /\ /\ /\ /\ /\ Odd row +--+--+--+--+--+--+--+
或者对于六边形：
__ __ __ __ / \__/ \__/ \__/ \__ Even row \__/ \__/ A\__/ \__/ Odd row / \__/ F\__/ B\__/ \__ Even row \__/ \__/ X\__/ \__/ Odd row / \__/ E\__/ C\__/ \__ Even row \__/ \__/ D\__/ \__/ Odd row / \__/ \__/ \__/ \__ Even row \__/ \__/ \__/ \__/ Odd row
然后,您可以将数据存储为常规2D数组.奇数行向右偏移.5,你需要弄清楚X的邻居步骤：上面：A =(0,-2),右上：B =(1,-1),右下：C = (1,1),下：D =(0,2),左下：E =(0,左上：F =(0,-1)
如果你没有浪费一点内存,你也可以将其他列留空,并且邻居变得更简单：(0,(1,( 0,( – 1,1)
总结
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