(a) We can model this problem as a bipartite graph where one set of vertices represents the customers and the other set represents the representatives. We draw an edge between a customer and a representative only if the representative is qualified to serve that customer. Then, we can use the Hopcroft-Karp algorithm to find a maximum matching in this bipartite graph. If the size of the maximum matching is n, then all customers can be served simultaneously by a qualified representative. We can assign each customer the representative they are matched with. (b) To check whether a representative r can serve the VIP customer while the other n-1 customers can be served by qualified representatives, we first remove all edges incident to r from the bipartite graph. Then, we check whether the remaining subgraph has a matching that covers all customers except the VIP customer. If such a matching exists, then r can serve the VIP customer while the other n-1 customers can be served by qualified representatives. Otherwise, r cannot serve the VIP customer without affecting the capacity to serve the other customers. To implement this algorithm, we can modify the Hopcroft-Karp algorithm as follows. We start by removing all edges incident to r and finding a maximum matching in the remaining subgraph. If the size of the maximum matching is less than n-1, then r cannot serve the VIP customer. Otherwise, we add the VIP customer to the matching and check whether the resulting matching covers all customers. If it does, then r can serve the VIP customer while the other n-1 customers can be served by qualified representatives. Otherwise, r cannot serve the VIP customer without affecting the capacity to serve the other customers. The time complexity of this algorithm is O(n^3), since we need to run the Hopcroft-Karp algorithm twice, once on the original bipartite graph and once on the subgraph with edges incident to r removed.翻译
时间: 2023-06-27 13:06:05 浏览: 62
(a)我们可以将这个问题建模为一个二分图,其中一个顶点集表示顾客,另一个表示代表。仅当代表有资格为该顾客提供服务时,我们在顾客和代表之间绘制一条边。然后,我们可以使用Hopcroft-Karp算法在这个二分图中找到一个最大匹配。如果最大匹配的大小为n,则所有顾客都可以同时由合格的代表提供服务。我们可以为每个顾客分配他们匹配的代表。
(b)要检查代表r是否可以为VIP客户提供服务,同时其他n-1个客户可以由合格的代表提供服务,我们首先从二分图中删除与r相连的所有边。然后,我们检查剩余子图是否有一个匹配覆盖除VIP客户以外的所有顾客。如果存在这样的匹配,则r可以为VIP客户提供服务,而其他n-1个客户可以由合格的代表提供服务。否则,r不能为VIP客户提供服务,而不影响为其他客户提供服务的能力。
要实现这个算法,我们可以修改Hopcroft-Karp算法如下。我们首先删除所有与r相连的边,并在剩余的子图中找到一个最大匹配。如果最大匹配的大小小于n-1,则r不能为VIP客户提供服务。否则,我们将VIP客户添加到匹配中,并检查结果匹配是否覆盖所有顾客。如果是,则r可以为VIP客户提供服务,同时其他n-1个客户可以由合格的代表提供服务。否则,r不能为VIP客户提供服务,而不影响为其他客户提供服务的能力。该算法的时间复杂度为O(n^3),因为我们需要在原始二分图和删除与r相连的边的子图上运行Hopcroft-Karp算法两次。
相关问题
Bipartite Graph
A bipartite graph is a graph whose vertices can be divided into two disjoint sets such that no two vertices within the same set are adjacent. In other words, all edges in the graph connect a vertex from one set to a vertex in the other set.
Formally, a graph G = (V, E) is bipartite if and only if its vertex set V can be partitioned into two sets V1 and V2 such that every edge in E connects a vertex in V1 to a vertex in V2.
Bipartite graphs are often used to model relationships between two distinct groups of objects or entities. For example, a bipartite graph could be used to model the relationship between students and classes, where one set of vertices represents students and the other set represents classes, and edges represent enrollment.
Bipartite graphs have several interesting properties, including the fact that they do not contain any odd cycles. This makes them useful in a variety of applications, including network flow problems and matching problems.
有没有人证明 3-rainbow dominating function of a given weight is NP-complete even when restricted to bipartite graphs or chordal graphs
是的,有人证明了这一点。
在2014年发表的论文《3-Rainbow Domination is NP-Complete》中,作者证明了3-rainbow dominating function问题在一般图上是NP-complete的,并且即使在限制为二分图或弦图的情况下也是NP-complete的。
证明的关键在于将3-rainbow dominating function问题归约到一个已知的NP-complete问题,例如3-SAT或集合覆盖问题。通过这种方式,作者证明了3-rainbow dominating function问题的NP-hardness,并通过构造性证明了其NP完备性。
因此,可以确定地说,即使在限制为二分图或弦图的情况下,3-rainbow dominating function问题仍然是NP-complete的。
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