旅行商关于额外部分订购的问题

发布于 2024-11-17 09:42:09 字数 395 浏览 6 评论 0原文

我正在寻找此问题的名称或任何关于算法或源代码的线索:

示例:您想要找到访问 100 个最大城市的最佳路线美国(经典 TSP),但在您访问任何特定城市之前,您必须访问该城市所在州的首府。

示例:您正在收集几位教授的学生的许可单。你需要拜访每一位学生和每一位教授,但在你见过教授的所有学生之前,你不能拜访一位教授。

一些谷歌搜索发现了顺序排序问题或“SOP”,但没有太多文献让我确信这是一个被广泛接受的名称。

我不认为这些部分排序可以在经典的 TSP 中简单地通过选择在图中使用哪些边来表示(例如,您最初不能从纽约去芝加哥,但是一旦您访问了斯普林菲尔德,您可以< /em>) 或权重,但我可能是错的。

I am looking for a name for this problem or any leads on an algorithm or source code:

Example: You want to find the best route to visit the 100 largest cities in the US (classic TSP) but before you can visit any given city you must visit the capital of the state it is in.

Example: You are collecting permission slips from the students of several professors. You need to visit every student and every professor but you can't visit a professor until you have seen all of his students.

Some Googling turns up the sequential ordering problem or "SOP" but there is not so much literature that I am convinced that this is a widely accepted name.

I don't think these partial orderings can be represented within the classic TSP simply by choosing which edges to use in the graph (e.g. you can't initially go from New York to Chicago, but once you visit Springfield you can) or weights, but I may be wrong.

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硪扪都還晓 2024-11-24 09:42:09

顺序排序问题由 Escudero 于 1988 年在一篇题为“顺序排序问题的不精确算法”的论文中首次提出(该论文发表在《欧洲运筹学杂志》上) ,所以这是问题的原始名称。论文摘要如下:

给定有向 G= (N, A) 和
惩罚矩阵 C,顺序矩阵
订购问题(以下简称SOP)
包括找到排列
集合 N 中的节点,使得
最小化基于 C 的函数并且执行
不违反优先顺序
由集合 A 给出的关系。
强大的充分条件
SOP实例的不可行性是
嵌入 SOP 的程序中
预处理。请注意,它是其中之一
任何算法中的关键步骤
尝试解决SOP。通过删除
与优先级相关的约束
关系,SOP可以转换为
经典的不对称旅行
推销员问题(以下简称 ATSP)。
该算法获得(希望)
通过修改得到满意的解决方案
相关问题的最优解
分配问题(以下简称 AP)如果
这不是一个可行的序列
订购(以下简称 FSO)。新的
解决方案“修补”子路线(如果
任何)优先考虑补丁
链接成本为零
弧线。基于 AP 的下限
ATSP 的最优解被收紧
通过使用一些给出的程序
在[1]中。无论如何,本地搜索
改进初始FSO是
执行;它使用 3 和 4 变化
为基础的程序。计算型
一系列广泛案例的结果是
已报道。

埃斯库德罗和他的合作者发表了许多关于该主题的论文,并引用了更多内容。如果您正在查阅文献,搜索他的论文或引用本文的论文可能会对您有所帮助。

SOP 是经过充分研究的非对称旅行商问题的约束版本,因此许多有关 ATSP 的文献可能与之相关。

The Sequential Ordering Problem was first introduced by Escudero in 1988 in a paper entitled "An Inexact Algorithm for the Sequential Ordering Problem" (this appeared in the European Journal of Operational Research), so this is the original name for the problem. The abstract of the paper reads:

Given the directed G= (N, A) and the
penalty matrix C, the Sequential
Ordering Problem (hereafter, SOP)
consists of finding the permutation of
the nodes from the set N, such that it
minimizes a C-based function and does
not violate the precedence
relationships given by the set A.
Strong sufficient conditions for the
infeasibility of a SOP's instance are
imbedded in a procedure for the SOP's
preprocessing. Note that it is one of
the key steps in any algorithm that
attempts to solve SOP. By dropping the
constraints related to the precedence
relationships, SOP can be converted in
the classical Asymmetric Traveling
Salesman Problem (hereafter, ATSP).
The algorithm obtains (hopefully)
satisfactory solutions by modifying
the optimal solution to the related
Assignment Problem (hereafter, AP) if
it is not a Feasible Sequential
Ordering (hereafter, FSO). The new
solution 'patches' the subtours (if
any) giving preference to the patches
with zero reduced cost in the linking
arcs. The AP-based lower bound on the
optimal solution to ATSP is tightened
by using some of the procedures given
in [1]. In any case, a local search
for improving the initial FSO is
performed; it uses 3- and 4-change
based procedures. Computational
results on a broad set of cases are
reported.

Escudero and his collaborators have a number of papers on the subject, with references to even more. Searching for papers by him or that reference this paper may help you if you're looking through the literature.

SOP is a well-studied constrained version of the Asymmetric Travelling Salesman Problem, so much of the literature on ATSP may be related.

高冷爸爸 2024-11-24 09:42:09

您可以构建一个考虑订购要求的状态机,用您的权重注释转换,并解决旅行推销员的问题。除非你有更多的节点:2^(大写字母数量)乘以原始节点数量。

You could build a state machine that takes into account the ordering requirements, annotate the transitions with your weights, and solve the traveling salesman on that. Except you'd have a lot more nodes: 2^(number of capitals) times the original number of nodes.

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