Java 如何使图遍历算法适用于大型图？

我正在编写一个算法，在给定源和目标时遍历图形。算法应该遍历并在图中显示从源到目标的所有可能路径。我使用广度优先搜索来实现这一点，它在一个小图形中工作。但是，当提供一个巨大的图形（有1000多个顶点）时，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。我能知道怎么解决这个问题吗

我进行的一个简单的测试（小图）

我的广度优先搜索算法

// 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;
}

---

1000个节点并不是一个很大的图。我想象你的代码中有一个无限循环。让我们看看这里有一个无限循环的可能位置：

while（！Q.isEmpty（））

啊，是的，我没有看到任何东西阻止您向队列添加节点。队列应该只处理每个节点一次。您需要保留已添加到队列且不允许再次添加的节点列表

它不会很快停止，因为您从未停止向队列中添加内容，所以它永远不会为空

这一天的教训是：当你有一个循环只有在满足某个条件时才结束时，要加倍确保该条件能够满足

// 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;
}