我不太清楚你是如何执行游戏的，但你可以这么做:

Determine the eyes location relative position to the gate. i.e. Is it left above? Right below? Then move the eyes opposite one of the two directions (such as make it move left if it is right of the gate, and below the gate) and check if there are and walls preventing you from doing so. If there are walls preventing you from doing so then make it move opposite the other direction (for example, if the coordinates of the eyes relative to the pin is right north and it was currently moving left but there is a wall in the way make it move south. Remember to keep checking each time to move to keep checking where the eyes are in relative to the gate and check to see when there is no latitudinal coordinate. i.e. it is only above the gate. In the case it is only above the gate move down if there is a wall, move either left or right and keep doing this number 1 - 4 until the eyes are in the den. I've never seen a dead end in Pacman this code will not account for dead ends. Also, I have included a solution to when the eyes would "wobble" between a wall that spans across the origin in my pseudocode.

一些伪代码:

   x = getRelativeOppositeLatitudinalCoord()
   y
   origX = x
    while(eyesNotInPen())
       x = getRelativeOppositeLatitudinalCoordofGate()
       y = getRelativeOppositeLongitudinalCoordofGate()
       if (getRelativeOppositeLatitudinalCoordofGate() == 0 && move(y) == false/*assume zero is neither left or right of the the gate and false means wall is in the way */)
            while (move(y) == false)
                 move(origX)
                 x = getRelativeOppositeLatitudinalCoordofGate()
        else if (move(x) == false) {
            move(y)
    endWhile

我不太清楚你是如何执行游戏的，但你可以这么做:

Determine the eyes location relative position to the gate. i.e. Is it left above? Right below? Then move the eyes opposite one of the two directions (such as make it move left if it is right of the gate, and below the gate) and check if there are and walls preventing you from doing so. If there are walls preventing you from doing so then make it move opposite the other direction (for example, if the coordinates of the eyes relative to the pin is right north and it was currently moving left but there is a wall in the way make it move south. Remember to keep checking each time to move to keep checking where the eyes are in relative to the gate and check to see when there is no latitudinal coordinate. i.e. it is only above the gate. In the case it is only above the gate move down if there is a wall, move either left or right and keep doing this number 1 - 4 until the eyes are in the den. I've never seen a dead end in Pacman this code will not account for dead ends. Also, I have included a solution to when the eyes would "wobble" between a wall that spans across the origin in my pseudocode.

一些伪代码:

   x = getRelativeOppositeLatitudinalCoord()
   y
   origX = x
    while(eyesNotInPen())
       x = getRelativeOppositeLatitudinalCoordofGate()
       y = getRelativeOppositeLongitudinalCoordofGate()
       if (getRelativeOppositeLatitudinalCoordofGate() == 0 && move(y) == false/*assume zero is neither left or right of the the gate and false means wall is in the way */)
            while (move(y) == false)
                 move(origX)
                 x = getRelativeOppositeLatitudinalCoordofGate()
        else if (move(x) == false) {
            move(y)
    endWhile

知道吃豆人的路径是非随机的(例如，每个特定的关卡0-255,inky, blinky, pinky和clyde将在该关卡中工作完全相同的路径)。

我会选择这个，然后猜测有一些主路径围绕整个 迷宫是眼球物体的“返回路径”，当吃豆人吃掉幽灵时，它就在那里。

我认为你的解决方案是正确的，比这更简单，就是制作一个更“现实”的新版本，鬼魂的眼睛可以穿过墙壁=)

如果每个正方形都有一个到中心的距离值呢?这样，对于每个给定的正方形，你可以在所有可能的方向上得到相邻正方形的值。你选择最小值的正方形，然后移动到那个正方形。

数值将使用任何可用的算法预先计算出来。

下面是ammoQ的洪水填充想法的模拟和伪代码。

queue q
enqueue q, ghost_origin
set visited

while q has squares
   p <= dequeue q
   for each square s adjacent to p
      if ( s not in visited ) then
         add s to visited
         s.returndirection <= direction from s to p
         enqueue q, s
      end if
   next
 next

它的思想是宽度优先搜索，所以每次你遇到一个新的相邻正方形s，最好的路径是经过p。我相信是O(N)。