In the unsplittable capacitated vehicle routing problem, we are given a metric space with a vertex called depot and a set of vertices called terminals. Each terminal is associated with a positive demand between 0 and 1. The goal is to find a minimum length collection of tours starting and ending at the depot such that the demand of each terminal is covered by a single tour (i.e., the demand cannot be split), and the total demand of the terminals in each tour does not exceed the capacity of 1. Our main result is a polynomial-time $(2+\epsilon)$-approximation algorithm for this problem in the two-dimensional Euclidean plane, i.e., for the special case where the terminals and the depot are associated with points in the Euclidean plane and their distances are defined accordingly. This improves on recent work by Blauth, Traub, and Vygen [IPCO'21] and Friggstad, Mousavi, Rahgoshay, and Salavatipour [IPCO'22].

翻译：在无法解决的机动车辆路由问题中,我们拥有一个称为仓库的顶端和一套称为终点站的网格空间,每个终点站都与0和1之间的正需求有关,目标是找到从仓库开始和终点的旅游的最低长度集合,以便每个终点站的需求由一次旅行满足(即需求不能分开),每次旅行的终端总需求不超过1. 我们的主要结果是,在两维的欧几里德飞机上,即终点站和仓库与欧几里德飞机的点有关的特殊情况下,对该问题采用多元时间(2 ⁇ epsilon)$-约合算法,并据此界定其距离,Blauth、Traub和Vygen[IPCO'21]以及Friggstad、Mousavi、Rahgoshay和Salavatipour[IPCO22]最近的工作有了改进。