用直线段求平面上的有效区域
Finding Efficient Region in The Plane with Line segments
摘要：
设$\mathscr O$为$\mathbb{R}^2$中的一组$n$不相交障碍物，$\mathscr M$为运动物体。令$s$和$l$分别表示移动对象$\mathscr M$的起始点和最大路径长度。给定${R}^2$中的点$p$，我们说点$p$对于$\mathscr M$是可实现的，使得$π(s,p)\leq l$，其中$π(\cdot)$表示存在障碍物时的最短路径长度。一个是找到区域$\mathscr R$，使得对于任何点$p\in \mathbb{R}^2$，如果对于$\mathscr M$可实现，则为$p\in \mathscr R$ ;否则为$p\notin \mathscr R$。在本文中，我们限制我们的注意力线段障碍的情况下。为了解决这个问题，我们开发了三种算法。为了直观起见，我们首先提出了一个更简单的算法。其基本思想是将问题归结为计算一组圆形可见区域（CVR）的并集。该算法需要$O(n^3)$时间。通过分析其主要步骤，我们突破了它的瓶颈，使用最短路径映射（SPM）技术，以获得这些圈（事先不可用），产生一个$O(n^2\log n)$算法。由于上述发现，第三种算法也使用SPM技术。然而，它没有继续建造CVR。相反，它直接遍历SPM的每个区域来跟踪边界，最终算法获得$O(n\log n)$复杂度。
英文摘要：
Let $\mathscr O$ be a set of $n$ disjoint obstacles in $\mathbb{R}^2$, $\mathscr M$ be a moving object. Let $s$ and $l$ denote the starting point and maximum path length of the moving object $\mathscr M$, respectively. Given a point $p$ in ${R}^2$, we say the point $p$ is achievable for $\mathscr M$ such that $π(s,p)\leq l$, where $π(\cdot)$ denotes the shortest path length in the presence of obstacles. One is to find a region $\mathscr R$ such that, for any point $p\in \mathbb{R}^2$, if it is achievable for $\mathscr M$, then $p\in \mathscr R$; otherwise, $p\notin \mathscr R$. In this paper, we restrict our attention to the case of line-segment obstacles. To tackle this problem, we develop three algorithms. We first present a simpler-version algorithm for the sake of intuition. Its basic idea is to reduce our problem to computing the union of a set of circular visibility regions (CVRs). This algorithm takes $O(n^3)$ time. By analysing its dominant steps, we break through its bottleneck by using the short path map (SPM) technique to obtain those circles (unavailable beforehand), yielding an $O(n^2\log n)$ algorithm. Owing to the finding above, the third algorithm also uses the SPM technique. It however, does not continue to construct the CVRs. Instead, it directly traverses each region of the SPM to trace the boundaries, the final algorithm obtains $O(n\log n)$ complexity.
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