在游戏开始前保存地图上的节点(交叉点) 当怪物死亡时，取点(坐标)并找到 节点列表中最近的节点 计算从该节点到洞的所有路径 按长度取最短路径 将该点与最近节点之间的空间长度相加 绘制并在路径上移动

享受吧!

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。

最初的《吃豆人》并没有使用寻径或花哨的AI。它只是让玩家觉得游戏比实际更有深度，但实际上它是随机的。正如Ian Millington和John Funge在《ai Intelligence for Games》中所述。

Not sure if it's true or not, but it makes a lot of sense to me. Honestly, I don't see these behaviors that people are talking about. Red/Blinky for ex is not following the player at all times, as they say. Nobody seems to be consistently following the player, on purpose. The chance that they will follow you looks random to me. And it's just very tempting to see behavior in randomness, especially when the chances of getting chased are very high, with 4 enemies and very limited turning options, in a small space. At least in its initial implementation, the game was extremely simple. Check out the book, it's in one of the first chapters.

Dtb23的建议是在每个角落随机选择一个方向，最终你会发现怪物洞听起来非常低效。

然而，你可以利用它低效的“回家”算法，通过在游戏难度中引入更多变化来让游戏变得更有趣。你可以通过应用上面的方法，比如你的路径点或洪水填充来做到这一点，但这样做是非确定性的。所以在每个角落，你都可以生成一个随机数来决定是走最优路线，还是随机方向。

随着玩家不断推进关卡，你将减少玩家选择随机方向的可能性。这将在关卡速度，幽灵速度，吃药丸暂停等之外为整体难度关卡添加另一个杠杆。你有更多的时间放松，而鬼魂只是无害的眼睛，但随着你的进步，时间会越来越短。

这是一个寻径问题。有关流行的算法，请参见http://wiki.gamedev.net/index.php/A*。

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