机器人路径是否有净空c？
Does a robot path have clearance c?
摘要：
多边形障碍物之间的大多数路径规划问题要求找到避开障碍物并且相对于某些度量或度量的组合是最优的路径，例如间隙至少为$c$的$u$到$v$的最短路径，其中$u$和$v$是自由空间中的点，并且$c$是正常数。在实际应用中，例如紧急干预/疏散和医疗计划，专家建议了许多$u$到$v$的路径，问题是这些路径是否满足特定要求，例如给定的离障碍物的间隙。我们解决以下路径查询问题：给定平面中$m$不相交简单多边形的集合$S$，其具有总共$n$顶点，预处理它们，使得对于由正常数$c$和具有$k$顶点的简单多边形路径$π$组成的查询，从自由空间中的点$u$到点$v$，其中$k$远小于$n$，可以快速地确定$π$是否具有至少$c$的间隙（即，在$π$的距离$c$内没有多边形障碍物）。为此，我们展示了如何解决以下相关问题：给定一组$S$的$m$简单多边形在$\Re^{2}$，预处理$S$到一个数据结构，使多边形在$S$最接近查询线段$s$可以迅速报告。我们提出了一个$O(t \log n)$时间，$O(t)$空间预处理，$O((n / \sqrt{t}) \log ^{7/2} n)$查询时间解决这个问题，对任何$n ^{1 + ε} \leq t \leq n^{2}$。对于具有$k$段的路径，这导致$O((n k / \sqrt{t}) \log ^{7/2} n)$查询时间，这是对当$k$小时可以从现有计算几何方法导出的算法的显著改进。
英文摘要：
Most path planning problems among polygonal obstacles ask to find a path that avoids the obstacles and is optimal with respect to some measure or a combination of measures, for example an $u$-to-$v$ shortest path of clearance at least $c$, where $u$ and $v$ are points in the free space and $c$ is a positive constant. In practical applications, such as emergency interventions/evacuations and medical treatment planning, a number of $u$-to-$v$ paths are suggested by experts and the question is whether such paths satisfy specific requirements, such as a given clearance from the obstacles. We address the following path query problem: Given a set $S$ of $m$ disjoint simple polygons in the plane, with a total of $n$ vertices, preprocess them so that for a query consisting of a positive constant $c$ and a simple polygonal path $π$ with $k$ vertices, from a point $u$ to a point $v$ in free space, where $k$ is much smaller than $n$, one can quickly decide whether $π$ has clearance at least $c$ (that is, there is no polygonal obstacle within distance $c$ of $π$). To do so, we show how to solve the following related problem: Given a set $S$ of $m$ simple polygons in $\Re^{2}$, preprocess $S$ into a data structure so that the polygon in $S$ closest to a query line segment $s$ can be reported quickly. We present an $O(t \log n)$ time, $O(t)$ space preprocessing, $O((n / \sqrt{t}) \log ^{7/2} n)$ query time solution for this problem, for any $n ^{1 + ε} \leq t \leq n^{2}$. For a path with $k$ segments, this results in $O((n k / \sqrt{t}) \log ^{7/2} n)$ query time, which is a significant improvement over algorithms that can be derived from existing computational geometry methods when $k$ is small.
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