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From: Alejandro Aragón (alex_aragon_at_[hidden])
Date: 2006-03-10 16:06:43


Douglas Gregor wrote:
> On Mar 6, 2006, at 12:23 AM, Alejandro Aragón wrote:
>> Doug,
>>
>> I tried to follow your directions but I coundn't find any concrete
>> example of using the get(vertex_all,orig,v) function. I did something
>> else, but I think it is not working. Could you give me further
>> directions to code it your way?
>>
>> This is what I came up with:
>>
>> // minimum_spanning_graph
>> Graph* graph::minimum_spanning_graph(std::vector<Edge>& spanning_tree)
>> {
>> // create an empty graph
>> // the original graph has pointer gPtr
>> Graph* copy = new Graph();
>>
>> // iterate over all vertices in the original graph
>> Vertex v_new;
>> for (tie(vi, vi_end) = vertices(*gPtr); vi != vi_end; ++vi){
>> v_new = add_vertex(*copy);
>
> I think you can change this line to:
>
> v_new = add_vertex(get(vertex_all, *gPtr, *vi), *copy);
>
>
>> coordmap[v_new] = coordmap[*vi];
>> }
>>
>> // iterate over all the edges in the list produced by algorithm
>> bool inserted;
>> Edge e_new;
>> for (std::vector < Edge >::iterator ei = spanning_tree.begin();
>> ei != spanning_tree.end(); ++ei) {
>> tie(e_new,inserted) = add_edge(source(*ei,*gPtr),target
>> (*ei,*gPtr),*copy);
>
> And then change this line to:
>
> tie(e_new, inserted) = add_edge(source(*ei, *gPtr), target(*ei,
> *gPtr), get(edge_all, *gPtr, *ei), *copy);
>
>> weightmap[e_new] = weightmap[*ei];
>> anglemap[e_new] = anglemap[*ei];
>> }
>>
>> std::cout<<"Number of edges in original graph =
>> "<<num_edges(*gPtr)<<std::endl;
>> std::cout<<"Number of edges in minimum spanning graph
>> ="<<num_edges(*copy)<<std::endl;
>>
>> return copy;
>
> boost/graph/copy.hpp uses edge_all and vertex_all; you might be able
> to look there for more hints.
>
> Doug

It works perfectly! Thanks for the changes Doug. I would like to go
further with this. I'm not an expert in c++ programming but I will try
to make this funciton a template and I will provide a couple of
examples. Maybe I will see it implemented in the library some day! =)

Alejandro Aragon


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