1. Introduction

All graphs considered in this paper are undirected, finite, and simple. Let be a graph with vertex set and edge set . Unless stated otherwise, it is understood that and . For a graph , we also use and to denote the vertex set and edge set of , respectively. A graph is a subgraph of if and . We use to denote a subgraph of   induced by a subset of , that is, is the subgraph with vertex set in which two vertices are adjacent whenever they are adjacent in . For any vertex , the neighborhood of in is and the closed neighborhood of in is . The degree of a vertex in , denoted by , is the number of edges incident with . If , then is an isolated vertex of . A clique is a subset of pairwise adjacent vertices of . A maximal clique is a clique that is not a proper subset of any other clique. We use to denote the collection of all maximal cliques of . A clique is maximum if there is no clique of of larger cardinality. The clique number of , denoted by , is the cardinality of a maximum clique of . We use to denote the collection of all maximum cliques of .

A maximum-clique transversal set of a graph is a subset of intersecting all maximum cliques of . The maximum-clique transversal number of , denoted by , is the minimum cardinality of a maximum-clique transversal set of . The maximum-clique transversal set problem is to find a maximum-clique transversal set of of minimum cardinality. A maximum-clique independent set of is a collection of pairwise disjoint maximum cliques of . The maximum-clique independence number, denoted by , is the maximum cardinality of a maximum-clique independent set of . The maximum-clique independent set problem is to find a maximum-clique independent set of of maximum cardinality.

Maximum-clique transversal sets were introduced by Chang et al. in 2001 [1]. One of the main objectives for their research on maximum-clique transversal sets is the placement of transmitter towers for cellular telephones. Chang et al. stated a cellular telephone tower placement problem as the maximum-clique transversal set problem. They considered the problem and presented fixed parameter and approximation results for planar graphs. They also investigated the problem for some other graph classes such as -trees, strongly chordal graphs, graphs with few s, comparability graphs, and distance-hereditary graphs. Recently, Lee [2] introduced some variations of the maximum-clique transversal set problem and presented complexity results for them on some well-known classes of graphs.

Maximum-clique transversal and maximum-clique independent sets are closely related to clique transversal and clique independent sets on graphs. A clique transversal set of a graph is a subset of intersecting all maximal cliques of and a clique independent set of is a collection of pairwise disjoint maximal cliques of . The clique transversal number of , denoted by , is the minimum cardinality of a clique transversal set of . The clique independence number of , denoted by , is the maximum cardinality of a clique independent set of . The clique transversal (resp., independent) set problem is to find a clique transversal (resp., independent) set of of minimum (resp., maximum) cardinality. The clique transversal and clique independent set problems have been widely studied in [1, 3–20].

In this paper, we study the weighted version of the maximum-clique transversal set problem. Let be a function assigning to each vertex of a weight such that all arithmetic operations on vertex weights can be performed in time . We call a vertex-weight function and call a weighted graph. We let for any subset of and let be the weight of . The weighted maximum-clique transversal set problem is to find a maximum-clique transversal set of a weighted graph such that is minimized.

We present polynomial-time algorithms (most of them with linear running time) for the weighted maximum-clique transversal set problem on split graphs, balanced graphs, strongly chordal graphs, Helly circular-arc graphs, comparability graphs, distance-hereditary graphs, and graphs of bounded treewidth.

2. Split Graphs

In this section, we consider the weighted maximum-clique transversal set problem on split graphs.

Definition 2.1. A split graph is a graph , where the vertices of can be partitioned into an independent set and a clique .

Throughout this section, we use to denote a split graph with a vertex-weight function . Without loss of generality, we may assume that has no isolated vertices and that the vertices of have been partitioned into an independent set and a maximum clique . We give Algorithm 1 to solve the weighted maximum-clique transversal set problem for a split graph .

Algorithm 1: Finding a maximum-clique transversal set of a split graph of minimum weight.

Theorem 2.2. Algorithm 1 finds a maximum-clique transversal set of a split graph of minimum weight in time.

Proof. The theorem holds trivially if has only one maximum clique. We may assume that has more than one maximum clique. We show the correctness of Algorithm 1 as follows.
Initially, and for each vertex . At each iteration of Steps (5)–(12), the algorithm removes from an element with , and is decreased by the weight if is adjacent to . At the end of the last iteration of Steps (5)–(12), the set consists of all vertices with , and for each vertex .
For every maximum clique of other than , it has exactly one vertex in and vertices in . For a vertex with , is not a maximum clique of . Therefore, .
Assume that is a maximum-clique transversal set of of minimum weight. Let and be two vertices in . For every maximum clique of other than , it has exactly one vertex in and vertices in . It is clear that contains at least one vertex in and thus contains at most two vertices in .
Let be a vertex in such that . Let be a vertex in such that and let be a vertex in such that . We consider the following two cases.Case 1 (). It can be easily verified that the set is a maximum-clique transversal set of of minimum weight, and thus .Case 2 (). Let and let . If , then is adjacent to every vertex in . We have . Suppose that . For any vertex , the maximum clique does not contain the vertex . Since does not contain any vertex in , must be included in . In other words, the set is a subset of . Then, the set is a maximum-clique transversal set of and . Note that and . We have Since , the set is a maximum-clique transversal set of of minimum weight. Following the discussion above, the algorithm is correct.
Clearly, Step (1) and Steps (13)–(21) of Algorithm 1 can be done in time. Steps (2)–(4) and Steps (5)–(12) can be done in time. Hence, the running time of the algorithm is time.

3. Balanced Graphs

In this section, we consider the weighted maximum-clique transversal set problem on balanced graphs.

Let be a graph. Suppose that , , and . A clique matrix (resp., maximum-clique matrix) of , is the (0,1)-matrix whose entry is 1 if (resp., ), and 0 otherwise. A (0,1)-matrix is balanced if it does not contain the vertex-edge incidence matrix of an odd cycle as a submatrix, or equivalently, if it does not contain a square submatrix of odd order with exactly two ones per row and column. A (0, 1)-matrix is totally balanced if it does not contain the vertex-edge incidence matrix of a cycle as a submatrix [21]. A graph is balanced (resp., totally balanced) if a clique matrix of is balanced (resp., totally balanced).

Balanced graphs have been considered in [2, 8, 12, 15, 22]. It can be easily verified that if a clique matrix of a graph is balanced (resp., totally balanced), then all clique matrices of are balanced (resp., totally balanced). Note that a maximum-clique matrix of a graph is a submatrix of some clique matrix of . By definition, a submatrix of a balanced matrix is also balanced. We have the following lemma.

Lemma 3.1. If a graph is balanced, then any maximum-clique matrix of is balanced.

Let 1 (resp., 0) be a vector with 1’s (resp., 0’s) and let be a column vector. Let be a weighted graph with . Suppose that is a maximum-clique matrix of . The weighted maximum-clique transversal set problem for can be formulated as the following integer linear programming problem:

Fulkerson et al. proved the following important property of balanced matrices.

Theorem 3.2 (Fulkerson et al. [22]). If is a balanced matrix, then the polyhedra and have only integer extreme points.

Theorem 3.3. For any weighted balanced graph , the weighted maximum-clique transversal set problem can be solved in polynomial time.

Proof. Balanced graphs are a subclass of hereditary clique-Helly graphs [8]. Prisner [23] showed that no connected hereditary clique-Helly graphs has more maximal cliques than edges. Then, a connected balanced graph has maximal cliques. Since all maximal cliques of a hereditary clique-Helly graph can be enumerated in polynomial time by the algorithms in [24], all the maximum cliques can be extracted in polynomial time. Therefore, a maximum-clique matrix of can be computed in polynomial time.
Note that if the extreme points of the polyhedra defined by the linear relaxation of an integer linear programming problem are integers, then the optimal solution of the integer linear programming problem is equal to the optimal solution of its linear relaxation. It is wellknown that linear programming problems can be solved in polynomial time. Following Lemma 3.1 and Theorem 3.2, the weighted maximum-clique transversal set problem is polynomial-time solvable for balanced graphs.

4. Strongly Chordal Graphs

In this section, we consider the weighted maximum-clique transversal set problem for strongly chordal graphs.

Let be a graph. A vertex is simplicial if all vertices of form a clique. The ordering of the vertices of is a perfect elimination ordering of if for all , is a simplicial vertex of the subgraph of induced by . Let denote the closed neighborhood of in . Rose [25] showed the characterization that a graph is chordal if and only if it has a perfect elimination ordering. A perfect elimination ordering is called a strong elimination ordering if the following condition is satisfied.

For if and belong to in , then .

Farber [26] showed that a graph is strongly chordal if and only if it admits a strong elimination ordering. So far, the fastest algorithm to recognize a strongly chordal graph and give a strong elimination ordering takes [27] or time [28].

Definition 4.1. Let be a graph and . Let . The -incidence graph of , denoted by , is defined as follows. The vertex set of is where . In , (1) is an independent set, (2) two vertices of are adjacent if they are adjacent in , and (3) for , is adjacent to if in .

Definition 4.2. A dominating set of a graph is a subset of such that for every vertex . The domination number of , denoted by , is the minimum cardinality of a dominating set of . The weighted dominating set problem is to find a dominating set of a weighted graph such that is minimized.

Lemma 4.3. Let be a weighted graph. Let be a vertex-weight function of defined by if and if . A maximum-clique transversal set of of minimum weight is equivalent to a dominating set of of minimum weight.

Proof. Let be the minimum weight of a maximum-clique transversal set of and let be the minimum weight of a dominating set of .
Suppose that is a maximum-clique transversal set of of minimum weight. Then every maximum clique of has at least one vertex in . By the construction of , a vertex is adjacent to the vertices of a maximum clique of . Therefore, is a dominating set of . We have .
Conversely, we let be a dominating set of of minimum weight. Clearly, does not contain any vertex in . For any vertex , the set has a vertex in . By the construction of , is a maximum clique of . Therefore, is a maximum-clique transversal of . We have .
Following the discussion above, and thus the lemma holds.

Lemma 4.4. Let be a strongly chordal graph. Then is a strongly chordal graph, and a strong elimination ordering of can be obtained from a strong elimination ordering of in time.

Proof. A strongly chordal graph is chordal. It has at most maximal cliques, and all of its maximal cliques can be enumerated in time [29]. Then, all maximum cliques of a strongly chordal graph can be enumerated in time.
Let and . The vertex-clique incidence graph of , denoted by , is defined as follows. The vertex set of is . In , (1) is an independent set, (2) two vertices of are adjacent if they are adjacent in , and (3) for , is adjacent to if in . Let be a maximum subset of such that is a maximum clique of for each vertex . Note that and . Therefore, is isomorphic to the subgraph of induced by .
Guruswami and Rangan [17] showed that is a strongly chordal graph with edges and vertices, and that a strong elimination ordering of can be obtained from a given one for in time. Therefore, is also a strongly chordal graph and can be constructed from in time.
Suppose that and . Then, . Let be a strong elimination ordering of . Let be the ordering of the vertices in obtained by removing all the vertices in from the ordering . It can be easily verified that the ordering is also a strong elimination ordering of . Following the discussion above, the lemma holds.

Theorem 4.5. The weighted maximum-clique transversal set problem can be solved in time for a strongly chordal graph if a strong elimination ordering is given.

Proof. It follows from Lemmas 4.3 and 4.4 and the result that the weighted dominating set problem can be solved in time for a strongly chordal graph if a strong elimination ordering is given [30].

5. Helly Circular-Arc Graphs

In the section, we consider the weighted maximum-clique transversal set problem for Helly circular-arc graphs.

Definition 5.1. Let be a set. Let be a set of some positive integers. A collection of subsets of is said to satisfy the Helly property if and for all implies .

Let be a collection of nonempty sets. The intersection graph of is obtained by representing each set in as a vertex and connecting two vertices with an edge if and only if their corresponding sets intersect. A circular-arc model is a pair , where is a circle and is a collection of arcs of . If a graph is the intersection graph of , then is a circular-arc graph. If satisfies the Helly property, then is a Helly circular-arc graph and is called a Helly circular-arc model of . For an arc , let be the vertex of corresponding to . For , let . Let be a point of and let be the collection of arcs that contain . If is a maximal clique of , then is called a clique point. Suppose that and are distinct points of . If , then and are equivalent. A clique point representation of is a maximum set of nonequivalent clique points of .

Let be a circular-arc model of a Helly circular-arc graph . Let be a maximal clique of and let be a clique point representation of . Due to Helly property, the arcs corresponding to vertices in have a point on the circle in common. It is clear that a point of is a clique point if and only if is a maximal clique of . Then, we have .

If and are points of , we use to denote an arc of starting at and ending at in clockwise direction. For each arc , the points are called the extremes of . We also use and to denote the starting point and ending point of the arc , respectively. Without loss of generality, we assume that (1) all arcs of are open arcs, (2) no single arc entirely covers , and (3) no two extremes of distinct arcs of coincide.

Let and for . An intersection segment is the contiguous part of formed by two consecutive extremes and , where is the starting point of some arc and is the ending point of an arc in clockwise direction. A point inside an intersection segment is called an intersection point. There are at most intersection segments and every clique point is an intersection point [13].

Let be a weighted Helly circular-arc graph with a Helly circular-arc model . Lin et al. [31] proposed a linear-time algorithm that finds a clique point representation of . Based upon their algorithm, we have the following theorem.

Theorem 5.2. Let be a Helly circular-arc model of a Helly circular-arc graph . Let be a clique point representation of . Then, the clique number and can be computed in linear time.

Guruswami and Rangan [17] showed that a Helly circular-arc graph has at most maximal cliques. The following lemma can be easily verified according to the Helly property.

Lemma 5.3. Suppose that is a Helly circular-arc graph with a Helly circular-arc model and . Let be a vertex of . Then, there are at most maximal cliques of containing the vertex .

Lemma 5.4. Suppose that is a Helly circular-arc graph with a Helly circular-arc model .(1)The -incidence graph has edges and vertices.(2)The -incidence graph is a Helly circular-arc graph and a Helly circular-arc model of can be obtained from in time.

Proof. (1) By Definition 4.1, . Since is a Helly circular-arc graph, it has at most maximal cliques [17]. Then has vertices. Without loss of generality, we assume that the clique number . By Lemma 5.3 and the construction of , every vertex is adjacent to at most vertices in . Hence, the number of is .
(2) Let be a clique point representation of . The clique point representation can be constructed in linear time [31]. Following Theorem 5.2, the clique number and the arc sets can be computed in linear time. Then .
Let be a maximum subset of such that is a maximum clique of for . The set and the arc sets can be computed in linear time. Then .
Let and let be a set of arcs of such that each arc of contains exactly one clique point of and contains no extremes of arcs of . It follows that is a Helly circular-arc graph and is a Helly circular-arc graph model for . Hence, is a Helly circular-arc graph, and a Helly circular-arc model of can be obtained from in time.

Theorem 5.5. The weighted maximum-clique transversal set problem can be solved in time for a Helly circular-arc graph if a Helly circular-arc model is given.

Proof. It follows from Lemmas 4.3 and 5.4, and the result that the weighted dominating set problem can be solved in time for a circular-arc graph if a circular-arc model is given [32].

6. Comparability Graphs

A directed graph (or just digraph) consists of a nonempty finite set of vertices and a finite set of ordered pairs of distinct vertices called arrows. We call the vertex set and the arrow set of . We also use and to denote the vertex set and arrow set of , respectively. For an arrow , the first vertex is its tail and the second vertex is its head. We also say that the arrow leaves and enters . For a vertex of , we use the following notations: The sets , and are called the outneighborhood, inneighborhood, and neighborhood of , respectively. We call the vertices in , and the outneighbors, inneighbors, and neighbors of , respectively.

A directed walk in a digraph from vertex to vertex , or simply a directed -walk is a sequence of vertices such that , and is an arrow in for , where is called the length of this walk. A directed path is a directed walk in which no vertex is repeated. A directed -path is directed path starting at and ending at . A directed Hamiltonian path is a directed path that visits each vertex of exactly once. A directed cycle is a directed -walk in which no vertex is repeated except . Arrow set is a transitive relation on if for all , the following holds:  

Let be an undirected graph. Then the directed graph is an orientation of if for all , either or and for all , holds. If is a transitive relation on , then is a transitive orientation of . If there are no directed cycles in , then is an acyclic orientation of . Assume that is a directed graph. An undirected graph is the underlying graph of if for all , and for all , either