On Apr 5, 2005, at 2:38 AM, Markus Krosche wrote:
Hello together,
in my software I have to represent bipartite graphs, in my case petri nets (two disjoint sets - the places and the transitions - and only links between elements of the disjoint sets). I would like to use BGL for this purpose, but I did not find a "nice" solution for this problem.
I am quit new to BGL and the only way I see is representing the two types of vertices inside a vertex-property. A better way would be to have two types of vertex properties, but I think this is not possible.
Is there anyone who has a solution or an idea for this kind of problem?
Well, you're probably stuck with one of two solutions: 1) Have both types of vertices stored inside a vertex property. You could make this property a pointer to a base class (e.g., BaseVertex*), where the derived class is the data for either the first or second set, e.g., struct BaseVertex { virtual ~BaseVertex() {} // things in common with both sets of vertex bool isInFirstSet() const = 0; }; struct SetOneVertex : BaseVertex { // things for the first set }; struct SetTwoVertex : BaseVertex { // things for the second set }; Or, if you're feeling like trying out more Boost libraries, you could use a boost::variant<FirstSetData, SecondSetData>. 2) Have two separate *external* property maps, one for the first set and one for the second set. For instance, the property maps could be iterator_property_map instances built from these vectors: std::vector<FirstSetData> first_set_data_vec(num_vertices(g)); std::vector<SecondSetData> second_set_data_vec(num_vertices(g)); typedef property_map<vertex_index_t, Graph>::const_type VertexIndexMap; VertexIndexMap index_map = get(vertex_index, g); iterator_property_map<FirstSetData*, VertexIndexMap> first_set_data(&first_set_data_vec[0], index_map); iterator_property_map<SecondSetData*, VertexIndexMap> second_set_data(&second_set_data_vec[0], index_map); Just be careful to only reference first_set_data from vertices in the first set and second_set_data from vertices in the second set. Doug