如何在许多路径之间创建一条路线

时间:2018-07-22 05:15:04

标签: java algorithm routing

我正试图从起点(绿色)到终点(红色)建立一条路线,如下图所示。

enter image description here

我已经制定了一种随机算法来制作路线,
以及每个顶点具有连接顶点的列表。例如[a]连接到点(c,d)

path = list of points(x,y),
#path =10
s_point = get starting point(x,y)
end_Point = get end point(x,y)
for(int i=0;i<=#paths;i++){
    path.add(s_point)
    checked_points.add(s_point)
    next_point = s_point
    while(next_point.not_equals(end_point)){
        _point[] = get points connected to (next_point)
        while(_point[random()].not checked){
            path.add(_point[random()])
            next_point = _point[random()]
        }
    }
}
get_use_of_the_shortest_path()

所以,问题是, 如何将随机方法替换为应该将路径直接引导到终点的方法(考虑可用的不同路径)???或有关如何制作路线算法的任何信息。

ex2,

2 个答案:

答案 0 :(得分:1)

考虑到路径上没有权重(例如Google地图,其中包含有关其道路上的交通和速度限制的信息),我建议使用BFS算法,该算法可找到边缘最少的路线

Here是Java的BFS的出色实现,如Aakash Hasija所写:

import java.io.*;
import java.util.*;

// This class represents a directed graph using adjacency list
// representation
class Graph
{
    private int V;   // No. of vertices
    private LinkedList<Integer> adj[]; //Adjacency Lists

    // Constructor
    Graph(int v)
    {
        V = v;
        adj = new LinkedList[v];
        for (int i=0; i<v; ++i)
            adj[i] = new LinkedList();
    }

    // Function to add an edge into the graph
    void addEdge(int v,int w)
    {
        adj[v].add(w);
    }

    // prints BFS traversal from a given source s
    void BFS(int s)
    {
        // Mark all the vertices as not visited(By default
        // set as false)
        boolean visited[] = new boolean[V];

        // Create a queue for BFS
        LinkedList<Integer> queue = new LinkedList<Integer>();

        // Mark the current node as visited and enqueue it
        visited[s]=true;
        queue.add(s);

        while (queue.size() != 0)
        {
            // Dequeue a vertex from queue and print it
            s = queue.poll();
            System.out.print(s+" ");

            // Get all adjacent vertices of the dequeued vertex s
            // If a adjacent has not been visited, then mark it
            // visited and enqueue it
            Iterator<Integer> i = adj[s].listIterator();
            while (i.hasNext())
            {
                int n = i.next();
                if (!visited[n])
                {
                    visited[n] = true;
                    queue.add(n);
                }
            }
        }
    }

    // Driver method to
    public static void main(String args[])
    {
        Graph g = new Graph(4);

        g.addEdge(0, 1);
        g.addEdge(0, 2);
        g.addEdge(1, 2);
        g.addEdge(2, 0);
        g.addEdge(2, 3);
        g.addEdge(3, 3);

        System.out.println("Following is Breadth First Traversal "+
                           "(starting from vertex 2)");

        g.BFS(2);
    }
}
// This code is contributed by Aakash Hasija

答案 1 :(得分:0)

您可以将距目标的曼哈顿距离用作启发式方法,以指导达到目标的路径。选择更接近的值。