确定远离其他Point的最远点

Question

我正在创建一个简单的游戏，在一些游戏计算机操舵玩家身上使用一些简单的AI实现。

我有一个Point列表，代表玩家的可能动作。我需要编写一种方法，将玩家移动到距离该列表中可能的敌人最远的Point。我用图片说明了它：

数字代表List

中的Points poistion

我想要的是让玩家（4）移动到距离任何敌人最远的位置2或6的Point。如果有一个敌人通过迭代列表并使用distance() Point方法确定哪个点距离最远，我就设法解决了这个问题。但即使网格中有几个敌人，代码也必须正常工作。

Answer 1

嗯，你怎么样呢：

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.

早期出局有很多潜力：

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

[edit]：如果你正在使用上面的网格，你可以

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.