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我正在创建一个简单的游戏，其中一些游戏计算机引导玩家使用一些简单的 AI 实现。

我有一个Point代表玩家可能移动的列表。我需要编写一个方法，将玩家移动到Point距离该列表中可能的敌人最远的地方。我用图片说明了它：

数字代表列表中的点位置

我想要的是让玩家（4）移动到Point距离任何敌人最远的位置 2 或 6。如果有一个敌人，我已经设法解决了这个问题，方法是迭代列表并使用distance()方法Point确定最远的点。但即使网格中有多个敌人，代码也必须工作。

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1 回答 1

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嗯，你反过来怎么样：

1. Iterate over each point.
2. Find out how close it is to its closest enemy.
3. Choose the point that is furthest from its closest enemy.

早期出局的潜力很大：

Within the loop store the currently furthest point. 
If you are then inspecting another point and find out
it has a closer enemy, you can immediately skip to the
next point

[编辑]：如果您使用上述网格，您也可以

1. Check if there's an enemy on the currently processed 
   point *before* iterating through other enemies. That way
   you can exclude it as early as possible.

2. If it's a densely populated grid, consider doing a breadth-first
   flood-fill starting at the current point. That might find the closest
   enemy much faster than iterating though all of them.
于 2012-07-18T09:31:27.230 回答