软件工程上机实验，包括四个问题，现已经完成四个

In a box bounded by [-1, 1], given m balloons(they cannot overlap) with variable radio r and position mu, find the optimal value of r and mu which maximize sum r^2

Solution:

将正方形面积等分成100*100个小方格，用一个链表记录小方格的交点，一个链表记录已经找出的圆。每次遍历这些交点，找出现有条件下的最大圆，将其加入已经找出的圆的链表。同时将圆内的点从交点链表中删除。直到找出所给数目的圆为止。

那么如何找出现有条件下的最大圆呢。最大圆一定是与圆或者是四条边想切的，因此只需求出当前点与四条边、所有已经找出的圆的距离，取最小值即可。

In a box bounded by [-1, 1], given m balloons(they cannot overlap) with variable radio r and position mu. And some tiny blocks are in the box at given position d;balloons cannot overlap with these blocks. find the optimal value of r and mu which maximizes sum $r^2

Solution:

相比较第一题来说，本题中增加了一些不可覆盖的点，那么我们可以将这些点作为半径为0的圆存入已经找出的圆链表中。在绘图时将这些圆删除即可。

###Q3 In a box(3D) bounded by [-1, 1], given m balloons(they cannot overlap) with variable radio r and position mu. And some tiny blocks are in the box at given position d;balloons cannot overlap with these blocks. find the optimal value of r and mu which maximizes sum $r^2

Solution:

本题就是将2D变成了3D，没有什么特别之处。球体的图形化展示使用了JavaFx。 ###Q4 Update to multi-process/thread/core implementation.In a box(3D) bounded by [-1, 1], given m balloons(they cannot overlap) with variable radio r and position mu. And some tiny blocks are in the box at given position d;balloons cannot overlap with these blocks. find the optimal value of r and mu which maximizes sum $r^2

Solution:

多线程优化方案：在计算当前情况下的所有点中的最大半径时，将所有的点划分成四部分，用四个线程来计算。但是使用多线程的方案由于线程的创建和上下文切换耗时比较高，效率反而没有单线程高。