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.

我的方法有点内存密集型(从《吃豆人》时代的角度来看)，但你只需要计算一次，它适用于任何关卡设计(包括跳跃)。

一次标记节点

当你第一次加载一个关卡时，将所有怪物巢穴节点标记为0(代表与巢穴的距离)。继续向外标记已连接的节点1，连接到它们的节点2，依此类推，直到所有节点都被标记。(注意:如果巢穴有多个入口，这也是有效的)

我假设您已经有了表示每个节点和到它们的邻居的连接的对象。伪代码可能看起来像这样:

public void fillMap(List<Node> nodes) { // call passing lairNodes
    int i = 0;

    while(nodes.count > 0) {
        // Label with distance from lair
        nodes.labelAll(i++);

        // Find connected unlabelled nodes
        nodes = nodes
            .flatMap(n -> n.neighbours)
            .filter(!n.isDistanceAssigned());
    }
}

眼睛移动到距离标签最小的邻居

一旦所有节点都标记好了，路由眼睛就变得很简单了……只需要选择距离标签最小的相邻节点(注意:如果多个节点的距离相等，那么选择哪个节点并不重要)。伪代码:

public Node moveEyes(final Node current) {
    return current.neighbours.min((n1, n2) -> n1.distance - n2.distance);
}

全标记示例

任何简单、可维护、可靠、性能足够好的解决方案都是好的解决方案。听起来你已经找到了一个很好的解决办法。

寻径解决方案可能比当前解决方案更复杂，因此更可能需要调试。它可能也会变慢。

在我看来，如果它没坏，就不要修。

EDIT

在我看来，如果迷宫是固定的，那么你当前的解决方案就是好的/优雅的代码。不要错误地把“好”或“优雅”等同于“聪明”。简单的代码也可以是“好的”和“优雅的”。

如果你有可配置的迷宫级别，那么也许你应该在最初配置迷宫时进行寻路。最简单的方法就是让迷宫设计师亲自动手。如果你有无数个迷宫，我才会费心自动化这个……或者用户可以自行设计。

(另外:如果路线是手工配置的，那么迷宫设计师可以通过使用次优路线来让关卡变得更有趣……)

我用这种方法解决了一般关卡的这个问题:在关卡开始前，我从怪物洞中进行某种“洪水填充”;迷宫中除了墙之外的每一块瓦都有一个数字，表示它离洞有多远。所以当眼睛盯着一个距离为68的瓦片时，他们会看哪个相邻的瓦片距离为67;那就这么办吧。

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=)