我使用 JGraphT 構建了下圖 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);但每當我嘗試打電話時：shortestPath.getPath(node1, node3);我得到一個空數組，這意味著沒有連接。我知道這可能與邊緣的方向有關，因為A->B->C工作正常。有沒有辦法找到路徑，而不管A和C之間的邊緣的方向如何？
1 回答
鴻蒙傳說
TA貢獻1865條經驗 獲得超7個贊
您可以使用無向圖類來實現此目的。
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);
注意：您可能希望使用 與 a 相對，除非您確實需要多個邊/自循環。SimpleDirectedGraphDirectedPseudograph
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