我发现在《吃豆人》中有很多关于幽灵AI的参考,但没有一个提到当幽灵被《吃豆人》吃掉后,眼睛是如何找到中央幽灵洞的。
在我的实现中,我实现了一个简单但糟糕的解决方案。我只是在每个角落都用硬编码标明了应该往哪个方向走。
有没有更好的/最好的解决办法?也许是适用于不同关卡设计的通用设计?
我发现在《吃豆人》中有很多关于幽灵AI的参考,但没有一个提到当幽灵被《吃豆人》吃掉后,眼睛是如何找到中央幽灵洞的。
在我的实现中,我实现了一个简单但糟糕的解决方案。我只是在每个角落都用硬编码标明了应该往哪个方向走。
有没有更好的/最好的解决办法?也许是适用于不同关卡设计的通用设计?
当前回答
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
其他回答
我建议幽灵存储他从洞到吃豆人的路径。所以一旦鬼魂死了,他就可以沿着这条存储路径向相反的方向移动。
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.
假设你已经有了追逐吃豆人所需的逻辑,为什么不重用它呢?只要改变目标。这似乎比尝试使用完全相同的逻辑创建一个全新的例程要少得多。
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。
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*。