使用Dijkstra检测多个最短路径 [英] Using Dijkstra to detect multiple shortest paths

问题描述

给定一个加权有向图，如何修改Dijkstra算法以测试给定节点对之间是否存在多个最低成本的路径？

Given a weighted directed graph, how can the Dijkstra algorithm be modified to test for the presence of multiple lowest-cost paths between a given pair of nodes?

我的当前算法如下：（贷给魏斯）

My current algorithm is as follows: (credit to Weiss)

/**
 * Single-source weighted shortest-path algorithm. (Dijkstra) 
 * using priority queues based on the binary heap
 */
public void dijkstra( String startName )
{
    PriorityQueue<Path> pq = new PriorityQueue<Path>( );

    Vertex start = vertexMap.get( startName );
    if( start == null )
        throw new NoSuchElementException( "Start vertex not found" );

    clearAll( );
    pq.add( new Path( start, 0 ) ); start.dist = 0;

    int nodesSeen = 0;
    while( !pq.isEmpty( ) && nodesSeen < vertexMap.size( ) )
    {
        Path vrec = pq.remove( );
        Vertex v = vrec.dest;
        if( v.scratch != 0 )  // already processed v
            continue;

        v.scratch = 1;
        nodesSeen++;

        for( Edge e : v.adj )
        {
            Vertex w = e.dest;
            double cvw = e.cost;

            if( cvw < 0 )
                throw new GraphException( "Graph has negative edges" );

            if( w.dist > v.dist + cvw )
            {
                w.dist = v.dist +cvw;
                w.prev = v;
                pq.add( new Path( w, w.dist ) );
            }
        }
    }
}

推荐答案

替换字段 prev ，链接到具有集合 prevs 的先前顶点，并稍微更改代码：

Replace field prev, the link to previous vertex with a collection prevs, and change the code slightly:

...

        if( w.dist >= v.dist + cvw ) {
            if ( w.dist > v.dist + cvw ) {
                w.dist = v.dist +cvw;
                w.prevs.clear();
            }
            w.prevs.add(v);
            pq.add( new Path( w, w.dist ) );
        }

...

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