找到有向图C ++的最短路径

Question

Over the last week, I have implemented a Digraph by parsing an input file. The graph is guaranteed to have no cycles. I have successfully created the graph, used methods to return the number of vertices and edges, and performed a topological sort of the graph. The graph is composed of different major courses and their prereqs. Here is my graph setup:

class vertex{
public:
    typedef std::pair<int, vertex*> ve;     
    std::vector<ve> adjacency;              
    std::string course;                     
    vertex(std::string c){
        course = c;
    }
};

class Digraph{
public:
    typedef std::map<std::string, vertex *> vmap;           
    vmap work;
    typedef std::unordered_set<vertex*> marksSet;           
    marksSet marks;
    typedef std::deque<vertex*> stack;                      
    stack topo;
    void dfs(vertex* vcur);                                 
    void addVertex(std::string&);                           
    void addEdge(std::string& from, std::string& to, int cost);     
    int getNumVertices();                                   
    int getNumEdges();                                      
    void getTopoSort();                                     

};

The implementation

//function to add vertex's to the graph
void Digraph::addVertex(std::string& course){
    vmap::iterator iter = work.begin();
    iter = work.find(course);
    if(iter == work.end()){
        vertex *v;
        v = new vertex(course);
        work[course] = v;
        return;
    }
}

//method to add edges to the graph
void Digraph::addEdge(std::string& from, std::string& to, int cost){
    vertex *f = (work.find(from)->second);
    vertex *t = (work.find(to)->second);
    std::pair<int, vertex *> edge = std::make_pair(cost, t);
    f->adjacency.push_back(edge);
}

//method to return the number of vertices in the graph
int Digraph::getNumVertices(){
    return work.size();
}

//method to return the number of edges in the graph
int Digraph::getNumEdges(){
    int count = 0;
    for (const auto & v : work) {
         count += v.second->adjacency.size();
     }
     return count;
}

//recursive function used by the topological sort method
void Digraph::dfs(vertex* vcur) {
  marks.insert(vcur);
  for (const auto & adj : vcur->adjacency) {
    vertex* suc = adj.second;
    if (marks.find(suc) == marks.end()) {
      this->dfs(suc);
    } 
  }
  topo.push_front(vcur);
}

//method to calculate and print out a topological sort of the graph
void Digraph::getTopoSort(){
    marks.clear();
    topo.clear();
    for (const auto & v : work) {
        if (marks.find(v.second) == marks.end()) {
            this->dfs(v.second);
        }
    }
    // Display it
   for (const auto v : topo) {
    std::cout << v->course << "\n";
  }
}

For the last part of my implementation, I have been trying to do 2 things. Find the shortest path from the first vertex to every other vertices, and also find the shortest path that visits every vertex and returns to the first one. I am completely lost on this implementation. I assumed from reading I need to use Dijkstra's algorithm to implement this. I have been trying for the last 3 days to no avail. Did i set up my digraph in a bad way to implement these steps? Any guidance is appreciated.

Answer 1

The fact that there are no cycles makes the problem much simpler. Finding the shortest paths and a minimal "grand tour" are O(n).

Implement Dijkstra and run it, without a "destination" node; just keep going until all nodes have been visited. Once every node has been marked (with its distance to the root), you can start at any node and follow the shortest (and only) path back to the root by always stepping to the only neighbor whose distance is less than this one. If you want, you can construct these paths quite easily as you go, and mark each node with the full path back to the root, but copying those paths can push the cost to O(n ² ) if you're not careful.

And once all the nodes are marked, you can construct a minimal grand tour. Start at the root; when you visit a node, iterate over its unvisited neighbors (ie all but the one you just came from), visiting each, then go back the one you came from. (I can put this with more mathematical rigor, or give an example, if you like.)