The $n$-vehicle exploration problem (NVEP) is a combinatorial search problem, which tries to find a permutation of $n$ vehicles to make one of the vehicles travel the maximal distance. Assuming the $n$ vehicles travel together, with some vehicles being used to refuel others, and at last all the vehicles return to the start point. NVEP has a fractional form of objective function, and its computational complexity of general case remains open. Given a directed graph $G$, if it has a Hamiltonian path, then the Hamiltonian path can be reduced in polynomial time to a single instance of NVEP, in which the traveling distance is at least $n$. The total length consists of $n$ parts, each of which is required to equal to $1$. We prove that $G$ has a hamiltonian path of total weight of $n$ if and only if the reduced NVEP instance has a sequence of length at least $n$ with each term equals to $1$. Therefore we show that Hamiltonian path $\leq_P$ NVEP, and consequently prove that NVEP is NP-complete.

翻译：n-车辆探索问题（NVEP）是一个组合搜索问题，旨在寻找n辆车的排列，使得其中一辆车行驶距离最大化。假设n辆车共同行驶，部分车辆用于为其他车辆加油，最终所有车辆返回起点。NVEP具有分数形式的目标函数，其一般情况的计算复杂度尚未解决。给定有向图G，若其存在哈密顿路径，则该哈密顿路径可在多项式时间内归约为NVEP的单个实例，其中行驶距离至少为n。总长度由n部分组成，每部分长度需等于1。我们证明，当且仅当归约后的NVEP实例存在一个每项均为1且长度至少为n的序列时，G才存在总权重为n的哈密顿路径。由此证明哈密顿路径≤P NVEP，从而证得NVEP是NP完全的。