BGL：邻接列表返回描述符的错误边缘属性

3 回答

• 0

  我相信你的错误来自代码库中的其他地方 .

  我把这个简单的代码放在一起测试边缘访问和类似图表上的顺序，一切都按预期工作 .

  如上所述，罪犯可以手动维护 edge_descriptors 或 vertex_descriptors . 或者也许你的路径矢量初始化或构造 .

  #include <iostream>
  #include <algorithm>
  #include <vector>
  
  #include <boost/graph/adjacency_list.hpp>
  #include <boost/graph/graph_traits.hpp>
  
  using namespace std;
  
  enum edge_t {A,B};
  
  struct MarkerVertex{
      std::string id;
  };
  
  struct RelationshipEdge{
      std::string id;
  };
  
  
  typedef boost::adjacency_list<boost::vecS, boost::vecS, boost::undirectedS,
      MarkerVertex, RelationshipEdge> CorrelationAdjacencyList;
  
  
  typedef boost::graph_traits<CorrelationAdjacencyList>::edge_descriptor   Edge;
  typedef boost::graph_traits<CorrelationAdjacencyList>::vertex_descriptor Vertex;
  
  typedef vector<Edge> PathType;
  
  
  void printPath(PathType &p, CorrelationAdjacencyList &g){
      cout << "print path" << endl;
  
      for(PathType::iterator pi = p.begin(); pi != p.end(); ++pi){
          Edge e = *pi;
  
          Vertex s = boost::source(e,g);
          Vertex t = boost::target(e,g);
  
          cout << g[s].id << "\t" 
               << g[e].id << "\t"
               << g[t].id << endl;
  
          bool isFound;
          Edge eForward;
          boost::tie(eForward,isFound) = boost::edge(s,t,g);
          cout << "forward  " << g[eForward].id << "\t" << isFound << endl;
  
          Edge eBackward;
          boost::tie(eBackward,isFound) = boost::edge(t,s,g);
          cout << "backward " << g[eBackward].id << "\t" << isFound << endl;
      }
  }
  
  
  int main(int argc, char* argv[]){
  
      CorrelationAdjacencyList graph;
  
      Vertex a = boost::add_vertex(graph); graph[a].id = "a";
      Vertex b = boost::add_vertex(graph); graph[b].id = "b";
      Vertex c = boost::add_vertex(graph); graph[c].id = "c";
  
      Edge e1 = boost::add_edge(a,b,graph).first; graph[e1].id = "e1";
      Edge e2 = boost::add_edge(b,c,graph).first; graph[e2].id = "e2";
      Edge e3 = boost::add_edge(c,a,graph).first; graph[e3].id = "e3";
  
      PathType path1,path2,path3;
  
      path1.push_back(e1);
      path1.push_back(e2);
      path1.push_back(e3);
  
      path2.push_back(e2);
      path2.push_back(e3);
      path2.push_back(e1);
  
      path3.push_back(e3);
      path3.push_back(e2);
      path3.push_back(e1);
  
      printPath(path1,graph);
      cout << endl;
      printPath(path2,graph);
      cout << endl;
      printPath(path3,graph);
  
  
      cin.get();
  }
  

  回复于 2024-05-04T12:38:34+08:00
• 2

  没有进一步的信息，我可以做出有根据的猜测，你使用 vecS 容器，迭代器/引用已经失效 .

  在插入/删除边/顶点时会发生这种情况 .

  回复于 2024-05-04T12:38:34+08:00
• 0

  这是我找到的解决方案，虽然我不相信我完全理解为什么原始方法没有 .

  我用上面的代码替换了

  for(PathType::iterator pathIterator = path.begin(); pathIterator != path.end(); ++pathIterator){
      edge_t edge = *pathIterator;
  
      int sourceID = m_graph[boost::source(edge, m_graph)].id;
      int destID = m_graph[boost::target(edge, m_graph)].id;
  
      int lowerID, higherID;
      if (sourceID < destID){
          lowerID = sourceID;
          higherID = destID;
      } else {
          lowerID = destID;
          higherID = sourceID;
      }
  
      edge_t edge2 = m_edgeDescriptorsByEndpoints.at(make_pair(lowerID, higherID));
      RelationshipEdge rEdge = m_graph[edge2];
  

  m_edgeDescriptorsByEndpoints 是边对描述符的顶点ID对的映射 .

  所以我正在获取边缘描述符，获取源顶点和目标顶点的ID，从 Map 中找到相应的边缘描述符，然后获取该边缘的属性 .

  不完全是一个令人满意的解决方案，但它确实有效，因为到目前为止我已经能够测试 .

  EDIT

  我按照@sehe的建议调用 boost::edge(u,v,g) 将调用替换为我的 Map ，所以代码现在如下：

  for(PathType::iterator pathIterator = path.begin(); pathIterator != path.end(); ++pathIterator){
              edge_t edge = *pathIterator;
  
              vert_t sourceV = boost::source(edge, m_graph);
              vert_t targetV = boost::target(edge, m_graph);
  
              pair<edge_t, bool> edgePair = boost::edge(sourceV, targetV, m_graph);
              if (!edgePair.second){
                  cerr << "Edge does not exist" << endl;
              }
              RelationshipEdge rEdge = m_graph[edgePair.first];
  

  这段代码仍会导致rEdge包含正确的属性，这看起来很奇怪，因为我认为查询边缘顶点的图形，然后查询图形中连接这些顶点的边缘总会给你完全相同的边后卫 .

  回复于 2024-05-04T12:38:34+08:00