Given a set P of n points in the plane, a unit-disk graph G_{r}(P) with respect to a radius r is an undirected graph whose vertex set is P such that an edge connects two points p, q \in P if the Euclidean distance between p and q is at most r. The length of any path in G_r(P) is the number of edges of the path. Given a value \lambda>0 and two points s and t of P, we consider the following reverse shortest path problem: finding the smallest r such that the shortest path length between s and t in G_r(P) is at most \lambda. It was known previously that the problem can be solved in O(n^{4/3} \log^3 n) time. In this paper, we present an algorithm of O(\lfloor \lambda \rfloor \cdot n \log n) time and another algorithm of O(n^{5/4} \log^2 n) time.
翻译:根据平面上设定的 n 点的设置 P, 单位- 磁盘图 G ⁇ r} (P) 对半径为 r 值而言, 单位- 磁盘图形 G ⁇ r} (P) 是一个未定向的图表, 其顶点设置为 P, 边缘连接两个点 p, q\ in P, 如果 p 和 q 之间的 Euclidean 距离最多为 r 。 G_r (P) 中的任何路径的长度是路径的边缘数 。 鉴于值 \ lambda > 0 和 2 rdot n 和 P, 我们考虑以下的逆向最短路径问题 : 找到最小的 r r, G_r (P) 和 t 之间的最短路径长度最多为\ lambda 。 先前知道 这个问题可以在 O( n @ 4/3} \ \ log3 n) 时间中解决 。 在本文中, 我们呈现 O ( l 底层\ lambda\ rd) 时间的算法 和 O (n) log2 n) 时间的另一种算算 。