PCG伪马泽算法？

Question

我使用photoshop创建了一组瓷砖：

我正在寻找一种方法来填充这些80x80瓷砖的屏幕，其中大多数区域是相连的。这是一个吃豆人风格游戏的游戏板，而不是一个有入口和出口的迷宫。

我唯一的目标是永远不要有一个面向墙壁的开口。

我目前通过从基于上一个瓷砖的子集中随机选择下一个瓷砖来实现这一点，因此如果最后一个瓷砖左侧有墙，那么下一个砖就不能左侧有开口。我还将拐角限制为较小的子集。这只适用于第一行。

我的问题是计算其余的行，这将不得不基于立即向左或立即向右的磁贴。我认为手工构建表格会变得很麻烦。（不是说努力工作有什么问题，而是低效的工作不好）。

这是我编写的代码，用于根据左侧的最新磁贴获取允许的下一个磁贴：

function getallowed(last_tile,x,y) as object

cselect={u : ["u","r","v","rx"]
            v : ["u","d","r","v","rx"]
            l : ["u","d","v","rx"]
            d : ["d","r","v","rx"]
            lx :["u","d","r","v","rx"]
            r : ["lx","ux","dx","h","c"]
            h : ["lx","ux","dx","l","h","c"]
            c : ["lx","ux","dx","l","h","c"]
            ux :["lx","ux","dx","l","h","c"]
            rx :["lx","ux","dx","l","h","c"]
            dx :["lx","ux","dx","l","h","c"]
            ul :["dx","br","ur","ux","lx","h","c"]
            ur :["ul","bl","rx","d","r","u","v"]
            bl :["ur","lx","ux","br","dx","c","h","l"]
            br :["ul","bl","rx","r","u","v","d"]}


toprow=["d","r","h","l","dx","ul","ur"]
'toprow=["d","dx","l","r"]
leftcol=["d","r","v","u","rx","ul","bl"]
rightcol=["d","l","v","u","lx","ur","br"]
bottomrow=["u","r","h","l","ux","bl","br"]



if x=0 then 
    if y=0 then
        ?"LEFTCOL AND TOPROW",x,y,last_tile
        selection=intersect(leftcol,toprow)
    else if y=8 then
        ?"LEFTCOL AND BOTTOMROW",x,y,last_tile
        selection=intersect(leftcol,bottomrow)
    else if y > 0 and y < 8 then
        ?"LEFTCOL AND FLOATING Y" ,x,y,last_tile
        selection=leftcol 'intersect(initsel,leftcol)
    end if
else if x=15 then 
    if y=0 then
        ?"RIGHTCOL AND TOPROW",x,y,last_tile
        selection=intersect(rightcol,toprow)
    else if y=8 then
        ?"RIGHTCOL AND BOTTOMROW",x,y,last_tile
        selection = intersect(rightcol,bottomrow)
    else if y>0 and y<8 then
        initsel = cselect[last_tile]
        ?"RIGHTCOL AND FLOATING Y",x,y,last_tile
        selection=intersect(initsel,rightcol)
    end if

    'x=1, y=0, last_tile= "ul"
else if x > 0 and x < 15 
    if y=0 then
        initsel = cselect[last_tile]
        ?"FLOATING X AND TOPROW",x,y,last_tile
        selection=intersect(initsel,toprow)
        'selection=toprow
    else if y=8 
        initsel = cselect[last_tile]
        ?"FLOATING X AND BOTTOM ROW",x,y,last_tile
        selection=intersect(initsel,bottomrow)
    else if y >0 and y < 8 then
        ?"NO CONDITIONS MATCHED RETURN FULL SET FOR LAST TILE",x,y,last_tile
        selection=cselect[last_tile]
    end if
end if
return selection
end function



function intersect(a, b)
    d = {}
    results = []

    for i = 0 to b.count()-1
        d[b[i]] = true
    end for
    for j = 0 to a.count()-1
        if d[a[j]] <> invalid then
            if d[a[j]] then
                results.push(a[j])
            end if
        end if
    end for
    return results
end function

使用我目前的方法，我将不得不为四个方向中的每一个创建一个表，并手动完成。

有人有其他建议或算法要建议吗？

Answer 1

如果你不想使用一般的迷宫生成算法，你可以尝试王瓦片或更高级的人字形王瓦片。