我目前已经实现了 Dijkstra 的算法,但是当我用这样的图测试我的算法时出现了问题:并尝试从 C 转到 B。我知道为什么它不起作用。但是我想知道如果有这样的图表,正常的实现是否会起作用? internal static Stack<string> Dijkstra(string sourcePoint, string targetPoint, Graph graph) { List<string> verticesStringList = graph.GetAllVertices(); Dictionary<string, Vertex> verticesDictionary = new Dictionary<string, Vertex>(); InitializeVerticesDictionary(sourcePoint, verticesStringList, verticesDictionary); while (verticesDictionary.Values.ToList().Any(x => x.IsVisited == false)) { KeyValuePair<string, Vertex> keyValuePair = verticesDictionary.Where(x => x.Value.IsVisited == false).ToList().Min(); string vertexKey = keyValuePair.Key; Vertex currentVertex = keyValuePair.Value; List<string> neighbourVertices = graph.GetNeighbourVerticesSorted(keyValuePair.Key); foreach (string neighbourVertexString in neighbourVertices) { Vertex neighbourVertex = verticesDictionary[neighbourVertexString]; int newDistanceFromStartVertex = currentVertex.ShortestDistanceFromTarget + graph.GetEdgeWeight(keyValuePair.Key, neighbourVertexString); if (newDistanceFromStartVertex < neighbourVertex.ShortestDistanceFromTarget) { verticesDictionary[neighbourVertexString].ShortestDistanceFromTarget = newDistanceFromStartVertex; verticesDictionary[neighbourVertexString].PreviousVertex = keyValuePair.Key; } } verticesDictionary[vertexKey].IsVisited = true; } return FormShortestPath(targetPoint, verticesDictionary); }更新:我改变了我的条件verticesDictionary.Values.ToList().Any(x => x.IsVisited == false && x.ShortestDistanceFromTarget != int.MaxValue),现在我没有得到我在评论中提到的溢出。
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郎朗坤
TA贡献1921条经验 获得超9个赞
IsVisited
这里有点误导,因为您实际上可以访问从源节点无法访问的节点。我会把它重命名为isProcessed
. 要检查您是否可以从源节点到达另一个节点,您需要检查它的距离是否为int.maxVal
。
为避免溢出,当 currentVertex.ShortestDistanceFromTarget 为 时不要迭代邻居int.maxVal
,因为它已经是源节点无法访问的节点。
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