![](/img/trans.png)
[英]in boost graph lib, how do I get a specific out-edge of a vertex without iterating over all the out-edges of that vertex?
[英]iterating over edges in boost reverse graph
我無法遍歷反向鄰接列表圖中的邊列表。 下面列出的最小代碼示例以及相應的clang 6 / gcc 7.3錯誤消息。 任何幫助將不勝感激!!!
#include <boost/config.hpp>
#include <algorithm>
#include <vector>
#include <utility>
#include <iostream>
#include <boost/graph/adjacency_list.hpp>
#include <boost/graph/reverse_graph.hpp>
#include <boost/graph/graph_utility.hpp>
using namespace boost;
struct vertexinfo {
int index{12};
};
struct edgeinfo {
int index{10};
};
using Graph = adjacency_list < vecS, vecS, bidirectionalS ,vertexinfo,edgeinfo>;
using ReversedGraph = reverse_graph<Graph>;
int main(){
Graph graph(5);
auto e1 = add_edge(0, 2, graph);
graph[e1.first].index = 1;
auto e2 = add_edge(1, 1, graph);
graph[e2.first].index = 2;
auto e3 = add_edge(1, 3, graph);
graph[e3.first].index = 3;
auto e4 = add_edge(1, 4, graph);
graph[e4.first].index = 4;
auto e5 = add_edge(2, 1, graph);
graph[e5.first].index = 5;
auto e6 = add_edge(2, 3, graph);
graph[e6.first].index = 6;
auto e7 = add_edge(2, 4, graph);
graph[e7.first].index = 7;
auto e8 = add_edge(3, 1, graph);
graph[e8.first].index = 8;
auto e9 = add_edge(3, 4, graph);
graph[e9.first].index = 9;
auto e10 = add_edge(4, 0, graph);
graph[e10.first].index = 10;
auto e11 = add_edge(4, 1, graph);
graph[e11.first].index = 11;
ReversedGraph reversed_graph(graph);
std::cout << "original graph:" << std::endl;
print_graph(graph, get(vertex_index, graph));
std::cout << std::endl << "reversed graph:" << std::endl;
print_graph(reversed_graph, get(vertex_index, graph));
auto range = edges(reversed_graph);
std::for_each(range.first,range.second,[&](const auto& item){
std::cout << reversed_graph[item].index << '\n';
});
return EXIT_SUCCESS;
}
錯誤消息包含在文件中
prog.cc:9: /opt/wandbox/boost-1.67.0/clang-6.0.0/include/boost/graph/reverse_graph.hpp:149:14: error: binding value of type 'const boost::adjacency_list<boost::vecS, boost::vecS, boost::bidirectionalS, vertexinfo, edgeinfo, boost::no_property, boost::listS>::edge_bundled' (aka 'const edgeinfo') to reference to type 'typename graph::detail::bundled_result<adjacency_list<vecS, vecS, bidirectionalS, vertexinfo, edgeinfo, no_property, listS>, typename detail::get_underlying_descriptor_from_reverse_descriptor<reverse_graph_edge_descriptor<edge_desc_impl<bidirectional_tag, unsigned long> > >::type>::type' (aka 'edgeinfo') drops 'const' qualifier
{ return m_g[detail::get_underlying_descriptor_from_reverse_descriptor<Descriptor>::convert(x)]; }
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ prog.cc:61:22: note: in instantiation of function template specialization 'boost::reverse_graph<boost::adjacency_list<boost::vecS, boost::vecS, boost::bidirectionalS, vertexinfo, edgeinfo, boost::no_property, boost::listS>, const boost::adjacency_list<boost::vecS, boost::vecS, boost::bidirectionalS, vertexinfo, edgeinfo, boost::no_property, boost::listS> &>::operator[]<boost::detail::reverse_graph_edge_descriptor<boost::detail::edge_desc_impl<boost::bidirectional_tag, unsigned long> > >' requested here
std::cout << reversed_graph[item].index << '\n';
^ prog.cc:60:10: note: in instantiation of function template specialization 'std::__1::for_each<boost::iterators::transform_iterator<boost::detail::reverse_graph_edge_descriptor_maker<boost::detail::edge_desc_impl<boost::bidirectional_tag, unsigned long> >, boost::detail::undirected_edge_iter<std::__1::__list_iterator<boost::list_edge<unsigned long, edgeinfo>, void *>, boost::detail::edge_desc_impl<boost::bidirectional_tag, unsigned long>, long>, boost::iterators::use_default, boost::iterators::use_default>, (lambda at prog.cc:60:44)>' requested here
std::for_each(range.first,range.second,[&](const auto& item){
從Boost Graph文檔中,反向圖形適配器(reverse_graph)的構造函數默認為GraphReference參數的const BidirectionalGraph&。 我必須將其更改為BidirectionalGraph&才能成功編譯代碼!
//using ReversedGraph = reverse_graph<Graph>;
using ReversedGraph = reverse_graph<Graph,Graph&>;
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.