rad*_*ead 3 java jgrapht shortest-path
我使用 JGraphT 构建了以下图 A->B<-C,如下所示:
DirectedPseudograph<Node, Edge> graph = new DirectedPseudograph<>(Edge.class);
DijkstraShortestPath<Node, Edge> shortestPath = new DijkstraShortestPath<Node, Edge>(graph);
Node bn1 = new Node("1", "A", null);
Node bn2 = new Node("2", "B", null);
Node bn3 = new Node("3", "C", null);
graph.addVertex(bn1);
graph.addVertex(bn2);
graph.addVertex(bn3);
Edge edge1 = new Edge("PART_OF");
Edge edge2 = new Edge("IS_A");
graph.addEdge(bn1, bn2, edge1);
graph.addEdge(bn3, bn2, edge2);
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但每当我尝试打电话时:
shortestPath.getPath(node1, node3);
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我得到一个空数组,这意味着没有连接。我知道这可能与边缘的方向有关,因为 A->B->C 工作正常。有没有办法无论A和C之间的边的方向如何都能找到路径?
您可以使用AsUndirectedGraph类来实现此目的。
Graph<Node, Edge> graph = new DirectedPseudograph<>(Edge.class);
Node bn1 = new Node("1", "A", null);
Node bn2 = new Node("2", "B", null);
Node bn3 = new Node("3", "C", null);
graph.addVertex(bn1);
graph.addVertex(bn2);
graph.addVertex(bn3);
Edge edge1 = new Edge("PART_OF");
Edge edge2 = new Edge("IS_A");
graph.addEdge(bn1, bn2, edge1);
graph.addEdge(bn3, bn2, edge2);
Graph<Node, Edge> undirGraph=new AsUndirectedGraph<>(graph);
ShortestPathAlgorithm<Node, Edge> shortestPath = new DijkstraShortestPath<Node, Edge>(undirGraph);
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注意:除非您确实需要多个边/自循环,否则您可能需要使用 aSimpleDirectedGraph而不是 a 。DirectedPseudograph
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