当不是所有节点都使用A *算法连接时，如何找到最短路径？

Question

Im trying to find the shortest path/best path from Bus Station A to Bus Station B using the A* algorithm where the bus route containing Station A and the bus route containing Station B might not have a common bus station.

(I would also like to change the cost of a move in such way so that i use as little buses as possible to reach my destination)

My BusRoute Class contains a linked list of bus stations,and each station has a name,latitude and longitude

The NODE Class i use to represent the states in A*:

public class Node
{
    double gCost;
    double hHeuristic;
    double f;//f=g+h
    Node parent;
    BusStation station;//contains lat,lng name
    String reached;//is either "Bus" or "Walking" , how i reached this node
    int busRouteId;//via which route i reached this station
}

gCost is the Havesine Distance between 2 stations

hHeuristic is the Havesine Distance between the current station and the goal

I keep expanding the node with the lowest f value from the priority queue util it has the goal station

This is the part where i generate all the possible transitions from the current node:

ArrayList<Node> posibleMoves=getPosibleMoves(current);
    for(Node p:posibleMoves){
        if(!visitedNodes.contains(p) ) {
            double predictedDistance = calculateHeuristic(p, goal);
            double cost;
            if (p.getReached().equals("Walking")) {
                cost=(10* havesineDistance(current.getLat(), current.getLng(), p.getLat(), p.getLng())) + current.getgCost();
            } else { //Reached using a bus
                cost=havesineDistance(current.getLat(), current.getLng(), p.getLat(), p.getLng()) + current.getgCost();
            }
            p.setgCost(cost);
            p.sethHeuristic(predictedDistance);
            p.setF(cost + predictedDistance);
            p.setParent(current);
            priorityQueue.add(p);
        }
    }
}

getPossibleMoves(current) finds all the bus routes containing this station and adds the next station from each found route to an arraylist.

After that it adds "Walking" Nodes to every other bus route that does not contain this station, trying to connect the routes togheter this way. (it finds the closest station from each route to the current station)

I set the cost of a Node reached by Walking to be 10 times the cost because a human is about 10 times slower than a bus.

After running the program i am not geting desired results, it either changes too many buses or it seems to walk for no reason. Also after checking the priority queue sometimes it doesnt poll the node with the lowest f value.

Is this a good way to aproach this problem? Am i missing something?

Answer 1

Most beginners in programming underestimate the difficulty of implementing pathplanning algorithm. A* is not a easy-going task, especially if is used with additional constraints. It must defined as a middle-size software project. I would guess that a vanilla A* algorithm needs 500 lines of code, while the extended variant of the OP with priority queue and heuristics needs 1000 lines of code. It is a good practice in software engineering to use UML charts for describing the interactions of the different classes. So my advice is, to download the dia software from the internet, and drawing around 10 classes in the chart for solving the A* algorithm. Programming a pathplanner without any UML chart and in the hope for doing it in one small method is some kind of blind flying.

Two exmple UML chart for pathplanning applications are here: [1] [2] They can't be used out-of-the box, but they are showing in which direction the project should go. They have around 10-15 boxes connected with edges which are describing a complex software project with lots of classes and methods. Otherwise, the amount of complexity can't be handled.