如何使图遍历算法与大图一起使用？

问题描述

我正在编写一种算法，以在给定源和目标时遍历图形。该算法应遍历并在图中显示从源到目的地的所有可能路径。我正在使用广度优先搜索来做到这一点，它在小图中起作用。但是，当提供了一个巨大的图形（具有超过1000个顶点）时，该程序似乎被冻结，并且在很长一段时间后不会显示任何输出。我可以知道如何解决这个问题吗？

我进行的一项简单测试（小图）

@H_502_5@    graph.addVertex("A",25);
    graph.addVertex("B",60);
    graph.addVertex("C",45);
    graph.addVertex("D",75);
    graph.addVertex("E",95);
    graph.addVertex("F",85);
    graph.addVertex("G",105);

    graph.addEdge("A","B","AB","");
    graph.addEdge("A","D","AD","C","AC","E","AE","");
    graph.addEdge("B","BE","");
    graph.addEdge("E","F","EF","G","EG","");
    graph.addEdge("D","DC","DF","");
    graph.addEdge("F","FG","");
    graph.addEdge("C","CF",""); 

    String str = graph.bfs("A","G");
    System.out.println(str);

我的广度优先搜索算法

@H_502_5@// SUBMODULE: bfs
// IMPORT: src (String),dest (String)
// EXPORT: T (String) 
public String bfs( String src,String dest )
{
    String T = "";
    DSAQueue Q = new DSAQueue();             // The queue will store linked list
    DSALinkedList path = null;                   // This is the list for queue
    DSALinkedList adj = null;                    // Adjacency for src

    adj = getAdjacent( src );

    // Make adjacent to linked list first and store in queue
    for( Object o : adj )
    {
        DSALinkedList ll = new DSALinkedList(); // Creating list for every iteration
        DSAGraphVertex vertex = (DSAGraphVertex) o;
        ll.insertLast( vertex );                // Every adjacent vertex is ele of list 
        Q.enqueue( ll );                        // Store the list to queue
    }

    // ASSERTION: We Iterate until Q is empty,Q will store all L[V]
    while( !Q.isEmpty() )
    {
        path = (DSALinkedList) Q.dequeue();          // Dequeue a linked list

        // Get the tail of list
        DSAGraphVertex v = (DSAGraphVertex) path.peekLast(); 

        // We found the complete list path when the tail is dest
        if( v.getLabel().equals(dest) )
        {
            T += src + " -> ";
            for( Object o : path )
            {
                DSAGraphVertex pathVertex = (DSAGraphVertex) o;
                T += pathVertex.getLabel() + " -> ";
            }
            T += "(END)\n";
        }
        else
        {
            // If v.getLabel() is not dest,we need to do bfs from adj of tail
            DSALinkedList tailAdj = v.getAdjacent();

            // Check the number of paths in this vertex
            if( v.getSize() == 1 ) 
            {
                /* If only one path found,simply traverse to the only one path */

                DSAGraphVertex headVertex = (DSAGraphVertex) tailAdj.peekFirst();    // head of the tailAdj
                path.insertLast( headVertex );                  
                Q.enqueue( path );
                headVertex = null;
            }
            else
            {
                /* If the vertex has more than one path,copy all the existing paths for 
                   every single connection,then insert the new node to each list */
                for( Object o : tailAdj )
                {
                    DSALinkedList newList = new DSALinkedList();
                    newList = copyList( path );                 // copy the existing

                    // Then add the new node into the copied list
                    DSAGraphVertex currVertex = (DSAGraphVertex) o; 
                    newList.insertLast( currVertex );

                    // The queue Now has a new list of paths 
                    Q.enqueue( newList );       
                    newList = null;
                }
            }
            tailAdj = null;
        }
        path = null;
    }
    return T;
}

解决方法

1000个节点在图形中不是很大。我想象您的代码中某处有一个无限循环。让我们在这里看到一个无限循环的可能位置：while( !Q.isEmpty() )

是的，我看不到阻止将节点添加到队列中的任何内容。队列仅应处理每个节点一次。您需要保留一个已添加到队列中并且不允许再次添加的节点的列表。

它不会很快停止，因为您永远不会停止向队列中添加内容，因此它永远不会为空。

当天的课程：只要您的循环只有在满足条件时才结束，所以请双重确保有可能满足该条件。