【问题标题】:Boost BGL BFS Find all unique paths from Source to TargetBoost BGL BFS 查找从源到目标的所有唯一路径
【发布时间】:2018-04-08 13:56:06
【问题描述】:

我正在使用 Boost BGL C++,我需要 Graph 来执行从源顶点到目标顶点的 BFS 并返回所有唯一路径。

现在,我想到了一种使用过滤图来获取包含从源到目标的路径的图的子集的方法,但我意识到它基本上不是过滤,因为过滤后的图包含访问但不是部分的顶点从源到目标的路径。有什么方法可以获取这些信息或其他方法更好吗?

参考代码:

boost::filtered_graph<DirectedGraph, boost::keep_all, std::function<bool(VertexDescr)>> Graph::getUniquePathsFromSource(VertexDescr source, VertexDescr target, DirectedGraph const & g)
{
    std::vector<double> distances(num_vertices(g));
    std::vector<boost::default_color_type> colormap(num_vertices(g));

    // Run BFS and record all distances from the source node
    breadth_first_search(g, source,
        visitor(make_bfs_visitor(boost::record_distances(distances.data(), boost::on_tree_edge())))
        .color_map(colormap.data())
    );

    for (auto vd : boost::make_iterator_range(vertices(g)))
        if (colormap.at(vd) == boost::default_color_type{})
            distances.at(vd) = -1;

    distances[source] = -2;

    boost::filtered_graph<DirectedGraph, boost::keep_all, std::function<bool(VertexDescr)>>
        fg(g, {}, [&](VertexDescr vd) { return distances[vd] != -1; });

    // Print edge list
    std::cout << "filtered out-edges:" << std::endl;
    std::cout << "Source Vertex: " << source << std::endl;

    auto ei = boost::edges(fg);

    typedef boost::property_map<DirectedGraph, boost::edge_weight_t>::type WeightMap;
    WeightMap weights = get(boost::edge_weight, fg);

    for (auto it = ei.first; it != ei.second; ++it)
    {
        if (source != boost::target(*it, g)) {
            std::cout << "Edge Probability " << *it << ": " << get(weights, *it) << std::endl;
        }
    }

    return fg;
}

输入(顶点1,顶点2,权重):

0 1 0.001
0 2 0.1
0 3 0.001
1 5 0.001
2 3 0.001
3 4 0.1
1 482 0.1
482 635 0.001
4 705 0.1
705 5 0.1
1 1491 0.01
1 1727 0.01
1 1765 0.01

输出(源 = 0,目标 = 5):

Source Vertex: 0
Edge Probability (0,1): 0.001
Edge Probability (0,2): 0.1
Edge Probability (0,3): 0.001
Edge Probability (1,5): 0.001
Edge Probability (1,482): 0.1
Edge Probability (1,1491): 0.01
Edge Probability (1,1727): 0.01
Edge Probability (1,1765): 0.01
Edge Probability (2,3): 0.001
Edge Probability (3,4): 0.1
Edge Probability (4,705): 0.1
Edge Probability (482,635): 0.001
Edge Probability (705,5): 0.1

预期输出:

0->1->5
0->3->4->705->5
0->2->3->4->705->5

【问题讨论】:

    标签: c++ boost graph


    【解决方案1】:

    我不会使用 BFS 算法,因为它使用颜色图来跟踪访问过的节点。但是,如果您想要 所有 个不同的路径,您将不想跳过已经访问过的节点(因为您可能会跳过替代路径)。

    相反,我会实现一个蛮力广度优先递归算法,它只访问所有相邻节点除非它们已经在当前路径中。

    所需的所有状态都是当前路径。

    这里更详细地解释了这个想法:https://www.quora.com/How-should-I-find-all-distinct-simple-paths-between-2-given-nodes-in-an-undirected-graph

    Live On Coliru

    #include <boost/graph/adjacency_list.hpp>
    #include <boost/graph/graph_utility.hpp> // print_graph
    using namespace boost;
    using Graph = adjacency_list<vecS, listS, directedS, property<vertex_index_t, int>, property<edge_weight_t, double> >;
    Graph read_graph();
    
    using Vertex = Graph::vertex_descriptor;
    using Path = std::vector<Vertex>;
    
    template <typename Report>
    void all_paths_helper(Vertex from, Vertex to, Graph const& g, Path& path, Report const& callback) {
        path.push_back(from);
    
        if (from == to) {
            callback(path);
        } else {
            for (auto out : make_iterator_range(out_edges(from, g))) {
                auto v = target(out, g);
                if (path.end() == std::find(path.begin(), path.end(), v)) {
                    all_paths_helper(v, to, g, path, callback);
                }
            }
        }
    
        path.pop_back();
    }
    
    template <typename Report>
    void all_paths(Vertex from, Vertex to, Graph const& g, Report const& callback) {
        Path state;
        all_paths_helper(from, to, g, state, callback);
    }
    
    int main() {
        auto g = read_graph();
        print_graph(g, std::cout);
    
        auto by_vertex_id = [&](int id) {
            return *find_if(vertices(g), [&](Vertex vd) { return id == get(vertex_index, g, vd); });
        };
    
        all_paths(by_vertex_id(0), by_vertex_id(5), g, [&](Path const& path) {
                std::cout << "Found path ";
                for (auto v : path)
                    std::cout << get(vertex_index, g, v) << " ";
                std::cout << "\n";
            });
        std::cout.flush();
    }
    
    // immaterial to the task, reading the graph
    Graph read_graph() {
        std::istringstream iss(R"(
            0 1 0.001
            0 2 0.1
            0 3 0.001
            1 5 0.001
            2 3 0.001
            3 4 0.1
            1 482 0.1
            482 635 0.001
            4 705 0.1
            705 5 0.1
            1 1491 0.01
            1 1727 0.01
            1 1765 0.01)");
    
        Graph g;
        auto vertex = [&,idx=std::map<int,Vertex>{}](int id) mutable {
            auto it = idx.find(id);
            if (it != idx.end())
                return it->second;
            return idx.emplace(id, add_vertex(id, g)).first->second;
        };
    
        for (std::string line; getline(iss, line);) {
            std::istringstream ls(line);
            int s,t; double w;
            if (ls >> s >> t >> w) {
                add_edge(vertex(s), vertex(t), w, g);
            } else {
                std::cerr << "Skipped invalid line '" << line << "'\n";
            }
        }
    
        return g;
    }
    

    哪些打印:

    1 --> 5 482 1491 1727 1765 
    0 --> 1 2 3 
    2 --> 3 
    3 --> 4 
    5 --> 
    4 --> 705 
    482 --> 635 
    635 --> 
    705 --> 5 
    1491 --> 
    1727 --> 
    1765 --> 
    Found path 0 1 5 
    Found path 0 2 3 4 705 5 
    Found path 0 3 4 705 5 
    

    【讨论】:

    • +1 为答案!我在没有auto by_vertex_id = [&amp;](int id) { return *find_if(vertices(g), [&amp;](Vertex vd) { return id == get(vertex_index, g, vd); }); }; 的情况下继续使用您的解决方案据我了解,这会检查 vertex_index 是否等于顶点描述符?
    • 你需要一些东西来找到源/目标顶点的描述符,对吧?如果你想更传统地拼写它(主要是更少的 lambda):pure c++11 甚至是pure c++03。 C++ 真的走了很长一段路
    • 啊,我明白了,我对 lambda 约定比较陌生,所以我还在学习它。谢谢!
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