C ++的boost ::图从向图得到父顶点 [英] c++ boost::graph get parent vertices from directed graph

问题描述

我有向图（通过从升压::图形库中的adjacency_graph实现），我试图找到某个顶点的父顶点。

在过去的（通过pygraph）我只是颠倒了有向图，然后做了邻居的搜索，但似乎有逆转的boost :: reverse_graph图令我有向图成图的双向，所以我不能使用该adjacent_vertices方法了。

有没有更好的方式来获取父顶点？

感谢。

下面是我当前的例子code：

 的#include＆LT;升压/图/ adjacency_list.hpp＆GT;
＃包括LT＆;升压/图/ reverse_graph.hpp＆GT;
＃包括LT＆;＆iostream的GT;TYPEDEF提振::的adjacency_list＆LT;提高::套，提振::血管内皮细胞，促进:: directedS＆GT;图形;
TYPEDEF提振:: reverse_graph＆LT;＆图表GT; Rgraph;
图表的typedef :: vertex_descriptor的顶点;诠释的main（）
{
    图图;
    顶点V0 =提振:: add_vertex（图）
    顶点V1 =升压:: add_vertex（图形）;
    顶点V2 =提振:: add_vertex（图）
    顶点V3 =提振:: add_vertex（图）
    顶点V4 =提振:: add_vertex（图）
    顶点V5 =提振:: add_vertex（图）
    顶点V6 =提振:: add_vertex（图）    升压::的add_edge（V0，V1，图表）;
    提高::的add_edge（V1，V2，图表）;
    提高::的add_edge（V2，V3，图表）;
    提高::的add_edge（V2，V4，图表）;
    提高::的add_edge（V3，V5，图表）;
    提高::的add_edge（V4，V5，图表）;
    提高::的add_edge（V5，V6，图表）;    图:: adjacency_iterator ibegin，IEND;
    对于（升压::领带（ibegin，IEND）=提振:: adjacent_vertices（V2，图表）！= ibegin IEND ++ ibegin）
    {
        性病::法院LT＆;＆LT; * ibegin＆LT;＆LT;的std :: ENDL;
    }    性病::法院LT＆;＆LT;的std :: ENDL＆LT;＆LT; ############# ############# RGRAPH＆LT;＆LT;的std :: ENDL＆LT;＆LT;的std :: ENDL;    Rgraph rgraph（图形）;
    Rgraph :: adjacency_iterator rbegin，分割;
    对于（升压::领带（rbegin，撕裂）=提振:: adjacent_vertices（V2，rgraph）; rbegin =撕裂;！++ rbegin）
    {
        性病::法院LT＆;＆LT; * rbegin＆LT;＆LT;的std :: ENDL;
    }
    性病::法院LT＆;＆LT;的std :: ENDL;    返回0;
}
 

解决方案

reverse_graph 要求调整图是模型<一个href=\"http://www.boost.org/libs/graph/doc/BidirectionalGraph.html\"><$c$c>BidirectionalGraph.如果你改变你的图表的typedef的boost ::的adjacency_list＆LT;提高::套，提振::血管内皮细胞，促进:: bidirectionalS＆GT;图; 你的程序编译，并给出结果：

  3
4############# ############# RGRAPH1
 

这是我相信的是你应该期望什么。

这不需要 reverse_graph （但仍需要 bidirectionalS ）是用另一种方式：

 走势:: out_edge_iterator out_begin，out_end;
对于（升压::领带（out_begin，out_end）= out_edges（V2，图形）; out_begin = out_end;！++ out_begin）
{
    性病::法院LT＆;＆LT;目标（* out_begin，图表）＆LT;＆LT;的std :: ENDL;
}
性病::法院LT＆;＆LT;的std :: ENDL;图:: in_edge_iterator in_begin，in_end;
对于（升压::领带（in_begin，in_end）= in_edges（V2，图形）; in_begin = in_end;！++ in_begin）
{
    性病::法院LT＆;＆LT;源（* in_begin，图表）＆LT;＆LT;的std :: ENDL;
}
性病::法院LT＆;＆LT;的std :: ENDL;
 

I have a directed graph (implemented via an adjacency_graph from the boost::graph library) and I'm trying to find the parent vertices of a certain vertex.

In the past (via pygraph) I have simply reversed the digraph, then done a neighbours search, but it appears that reversing the graph with boost::reverse_graph turns my digraph into a bidirectional graph, and therefore I can't use the adjacent_vertices method anymore.

Is there a better way to get the parent vertices?

Thanks.

Here's my current example code:

#include <boost/graph/adjacency_list.hpp>
#include <boost/graph/reverse_graph.hpp>
#include <iostream>

typedef boost::adjacency_list< boost::setS, boost::vecS, boost::directedS > Graph;
typedef boost::reverse_graph<Graph> Rgraph;
typedef Graph::vertex_descriptor Vertex;

int main()
{
    Graph graph;
    Vertex v0 = boost::add_vertex(graph);
    Vertex v1 = boost::add_vertex(graph);
    Vertex v2 = boost::add_vertex(graph);
    Vertex v3 = boost::add_vertex(graph);
    Vertex v4 = boost::add_vertex(graph);
    Vertex v5 = boost::add_vertex(graph);
    Vertex v6 = boost::add_vertex(graph);

    boost::add_edge(v0,v1,graph);
    boost::add_edge(v1,v2,graph);
    boost::add_edge(v2,v3,graph);
    boost::add_edge(v2,v4,graph);
    boost::add_edge(v3,v5,graph);
    boost::add_edge(v4,v5,graph);
    boost::add_edge(v5,v6,graph);

    Graph::adjacency_iterator ibegin, iend;
    for (boost::tie(ibegin, iend) = boost::adjacent_vertices(v2, graph); ibegin != iend; ++ibegin)
    {
        std::cout << *ibegin << std::endl;
    }

    std::cout << std::endl << "############# RGRAPH #############" << std::endl << std::endl;

    Rgraph rgraph(graph);
    Rgraph::adjacency_iterator rbegin, rend;
    for (boost::tie(rbegin, rend) = boost::adjacent_vertices(v2, rgraph); rbegin != rend; ++rbegin)
    {
        std::cout << *rbegin << std::endl;
    }
    std::cout << std::endl;

    return 0;
}

解决方案

reverse_graph requires that the adapted graph be a model of BidirectionalGraph. If you change your graph to typedef boost::adjacency_list< boost::setS, boost::vecS, boost::bidirectionalS > Graph; your program compiles and gives the result:

3
4

############# RGRAPH #############

1

that I believe is what you should expect.

Another way that does not require the reverse_graph (but still requires bidirectionalS) is to use:

Graph::out_edge_iterator out_begin, out_end;
for (boost::tie(out_begin, out_end) = out_edges(v2,graph); out_begin != out_end; ++out_begin)
{   
    std::cout << target(*out_begin,graph) << std::endl;
}
std::cout << std::endl;

Graph::in_edge_iterator in_begin, in_end;
for (boost::tie(in_begin, in_end) = in_edges(v2,graph); in_begin != in_end; ++in_begin)
{   
    std::cout << source(*in_begin,graph) << std::endl;
}
std::cout << std::endl;

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