Question

我正在尝试实现Dijkstra算法，该算法可以找到起始节点和结束节点之间的最短路径。在到达终端节点之前，有一些“必须通过”的中间节点（多于一个），例如2或3必须通过必须在到达终端节点之前通过的节点。

如果我必须通过一个节点，我找到的解决方案是找到从必须传递节点到目的地的两条不同路径，并从必须传递节点到启动节点。

我没有想法如何实现这样的算法。有什么建议吗？

感谢。

List<Node> closestPathFromOrigin = null;

double maxD = Double.POSITIVE_INFINITY;
double _distance = 0;
int temp1 = 0;
List<Node> referencePath = new ArrayList<>();
boolean check = false;
Node startNode = null;

public List<Node> recursion(ArrayList<Node> points, ArrayList<Node> intermediatePoints) {

    if (!check) {
        System.out.println("--- DATA ---");
        System.out.println("Intermediate points: " + intermediatePoints);
        System.out.println("points: " + points.get(0).lat + " " + points.get(1).lat);
        System.out.println("--Find the nearest intermediate point from the start point of driver--");
        startNode = points.get(0);
        System.out.println("Start point of driver: " + startNode.lat + " " + startNode.lon);
        for (int i = 0; i < intermediatePoints.size(); i++) {
            List<Node> _path = dijkstra(startNode, intermediatePoints.get(i));
            _distance = 0;
            for (int j = 0; j < _path.size() - 1; j++) {
                _distance += calculateDistance(_path.get(j), _path.get(j + 1));
            }
            if (_distance < maxD) {
                maxD = _distance;
                closestPathFromOrigin = _path;
                temp1 = i;
            }
        }
        System.out.println("NearestPoint from driver's origin: " + intermediatePoints.get(temp1));

        referencePath.addAll(closestPathFromOrigin);
        startNode = intermediatePoints.get(temp1);
        System.out.println("New StartNode: the nearestPoint from driver's origin: " + startNode.lat + " " + startNode.lon);
        check = true;
        intermediatePoints.remove(intermediatePoints.get(temp1));
        System.out.println("New Intermediate points: " + intermediatePoints);
        System.out.println("Intermediate points empty? No -> recursion, Yes -> stop");
        if (!intermediatePoints.isEmpty()) {
            System.out.println("Recursion!!! with new data of: intermediatePoints: " + intermediatePoints);
            recursion(points, intermediatePoints);
        } else {
            System.out.println("Stop");
            return referencePath;
        }
    } else {
        System.out.println("Recursion: startNode: " + startNode.lat + " " + startNode.lon);
        for (int i = 0; i < intermediatePoints.size(); i++) {
            if (intermediatePoints.size() > 1) {
                System.out.println("From the new start point to the next nearest intermediate points if more than one points");
                List<Node> _path = dijkstra(startNode, intermediatePoints.get(i));
                _distance = 0;
                for (int j = 0; j < _path.size() - 1; j++) {
                    _distance += calculateDistance(_path.get(j), _path.get(j + 1));
                }
                if (_distance < maxD) {
                    maxD = _distance;
                    closestPathFromOrigin = _path;
                    temp1 = i;
                }
                referencePath.addAll(closestPathFromOrigin);
                startNode = intermediatePoints.get(temp1);
                check = true;
                intermediatePoints.remove(intermediatePoints.get(temp1));
                if (!intermediatePoints.isEmpty()) {
                    recursion(points, intermediatePoints);
                } else {
                    return referencePath;
                }
            } else {
                System.out.println("From the new start point to the next nearest intermediate points if just one point");
                List<Node> _path = dijkstra(startNode, intermediatePoints.get(i));
                //Collections.reverse(_path);
                referencePath.addAll(_path);
            }
            if (i == intermediatePoints.size() - 1) {
                System.out.println("Last Entry in intermediate points - find path to destination: " + points.get(1).lat + " " + intermediatePoints.get(i));
                //List<Node> _path1 = dijkstra(points.get(1), intermediatePoints.get(i));
                List<Node> _path1 = dijkstra(intermediatePoints.get(i), points.get(1));

                Collections.reverse(_path1);
                referencePath.addAll(_path1);
               //  referencePath.addAll(_path2);
            }
        }
    }
    return referencePath;
}

Answer 1

这是旅行商问题的概括。 TSP出现的情况是所有顶点都是“必须通过”。

找到每对必须通过顶点之间的最短路径，从源到每个必须通过的顶点，以及从每个必须通过的顶点到接收器。然后使用着名的O（n 2 ^ n）动态编程算法为TSP找到满足约束条件的从源到接收器的最短路径;这里n将是两个加上必须通过顶点的数量。

Answer 2

通过查找必须包含节点和两个（结束和开始）节点之间的最短路径。表格图，然后运行最短路径算法（Dijkstra算法）。开始和结束节点将是相同的。

Answer 3

不幸的是这个问题被简化为TSP所以不要期望多项式解决方案，但如果没有中间节点那么小，那么你可以像下面这样快速地做到这一点： -

尝试每个节点序列都可以访问。

说你有s-＆gt; a-＆gt; b-＆gt; c-＆gt; d

然后使用dijkstra评估min（s，d）+ min（d，a）+ min（c，d）

最小距离的序列就是你的答案。

时间复杂度： O(k!*ElogV)其中k表示必须传递节点

Dijkstra的算法带有'必须通过'节点

3 个答案: