Question

目前我有一个应用程序可以根据Robert Sedgewick的代码计算加权图的最短路径。

我想要做的是计算路径的权重，路径的长度和已检查的顶点数。我已经成功找到了前两个，但找到已经检查过的顶点数量并没有真正起作用。

举一个我想做的例子：

我想知道在下图中红线检查了多少个方块： http://prntscr.com/9921cv

任何关于如何开始的线索都会非常受欢迎。

这就是我的代码目前的样子：

public class Dijkstra {

private double[] distTo;          // distTo[v] = distance  of shortest s->v path
private DirectedEdge[] edgeTo;    // edgeTo[v] = last edge on shortest s->v path
private IndexMinPQ<Double> pq;    // priority queue of vertices
private int length;               // length of a path


/**
 * Computes a shortest paths tree from <tt>s</tt> to every other vertex in
 * the edge-weighted digraph <tt>G</tt>.
 * @param G the edge-weighted digraph
 * @param s the source vertex
 * @throws IllegalArgumentException if an edge weight is negative
 * @throws IllegalArgumentException unless 0 &le; <tt>s</tt> &le; <tt>V</tt> - 1
 */
public Dijkstra(EdgeWeightedDigraph G, int s) {
    for (DirectedEdge e : G.edges()) {

        if (e.weight() < 0) {
            throw new IllegalArgumentException("edge " + e + " has negative weight");
        }
    }

    distTo = new double[G.V()];
    edgeTo = new DirectedEdge[G.V()];

    for (int v = 0; v < G.V(); v++) {
        distTo[v] = Double.POSITIVE_INFINITY;

    }
    distTo[s] = 0.0;

    // relax vertices in order of distance from s
    pq = new IndexMinPQ<Double>(G.V());
    pq.insert(s, distTo[s]);

    while (!pq.isEmpty()) {
        int v = pq.delMin();
        for (DirectedEdge e : G.adj(v))
            relax(e);

    }


    // check optimality conditions
    assert check(G, s);
}

// relax edge e and updat e pq if changed
private void relax(DirectedEdge e) {
    int v = e.from(), w = e.to();
    if (distTo[w] > distTo[v] + e.weight()) {
        distTo[w] = distTo[v] + e.weight();
        edgeTo[w] = e;
        if (pq.contains(w)) pq.decreaseKey(w, distTo[w]);
        else   {
            pq.insert(w, distTo[w]);                
        }            
    }

}

public int getLength() {
    return length;
}

public IndexMinPQ<Double> getPq() {
    return pq;
}

/**
 * Returns the length of a shortest path from the source vertex <tt>s</tt> to vertex <tt>v</tt>.
 * @param v the destination vertex
 * @return the length of a shortest path from the source vertex <tt>s</tt> to vertex <tt>v</tt>;
 *    <tt>Double.POSITIVE_INFINITY</tt> if no such path
 */
public double distTo(int v) {
    return distTo[v];
}

/**
 * Is there a path from the source vertex <tt>s</tt> to vertex <tt>v</tt>?
 * @param v the destination vertex
 * @return <tt>true</tt> if there is a path from the source vertex
 *    <tt>s</tt> to vertex <tt>v</tt>, and <tt>false</tt> otherwise
 */
public boolean hasPathTo(int v) {
    return distTo[v] < Double.POSITIVE_INFINITY;
}

/**
 * Returns a shortest path from the source vertex <tt>s</tt> to vertex <tt>v</tt>.
 * @param v the destination vertex
 * @return a shortest path from the source vertex <tt>s</tt> to vertex <tt>v</tt>
 *    as an iterable of edges, and <tt>null</tt> if no such path
 */
public Iterable<DirectedEdge> pathTo(int v) {
    if (!hasPathTo(v)) return null;
    Stack<DirectedEdge> path = new Stack<DirectedEdge>();
    for (DirectedEdge e = edgeTo[v]; e != null; e = edgeTo[e.from()]) {
        path.push(e);
        length++;
    }
    return path;
}


// check optimality conditions:
// (i) for all edges e:            distTo[e.to()] <= distTo[e.from()] + e.weight()
// (ii) for all edge e on the SPT: distTo[e.to()] == distTo[e.from()] + e.weight()
private boolean check(EdgeWeightedDigraph G, int s) {

    // check that edge weights are nonnegative
    for (DirectedEdge e : G.edges()) {
        if (e.weight() < 0) {
            System.err.println("negative edge weight detected");
            return false;
        }
    }

    // check that distTo[v] and edgeTo[v] are consistent
    if (distTo[s] != 0.0 || edgeTo[s] != null) {
        System.err.println("distTo[s] and edgeTo[s] inconsistent");
        return false;
    }
    for (int v = 0; v < G.V(); v++) {
        if (v == s) continue;
        if (edgeTo[v] == null && distTo[v] != Double.POSITIVE_INFINITY) {
            System.err.println("distTo[] and edgeTo[] inconsistent");
            return false;
        }
    }

    // check that all edges e = v->w satisfy distTo[w] <= distTo[v] + e.weight()
    for (int v = 0; v < G.V(); v++) {
        for (DirectedEdge e : G.adj(v)) {
            int w = e.to();
            if (distTo[v] + e.weight() < distTo[w]) {
                System.err.println("edge " + e + " not relaxed");
                return false;
            }
        }
    }

    // check that all edges e = v->w on SPT satisfy distTo[w] == distTo[v] + e.weight()
    for (int w = 0; w < G.V(); w++) {
        if (edgeTo[w] == null) continue;
        DirectedEdge e = edgeTo[w];
        int v = e.from();
        if (w != e.to()) return false;
        if (distTo[v] + e.weight() != distTo[w]) {
            System.err.println("edge " + e + " on shortest path not tight");
            return false;
        }
    }
    return true;
}
 }

Answer 1

找到已经检查过的顶点数量并没有真正起作用。

每次从队列中弹出一个顶点时，递增一个计数器（让我们称之为vertexCounter）：

int vertexCounter = 0;
while (!pq.isEmpty()) {
    int v = pq.delMin();
    vertexCounter++;
    for (DirectedEdge e : G.adj(v))
        relax(e);
}

顺便说一句，您应该改进所有变量和方法的名称。很难读懂你的代码。

Answer 2

您可以通过向类中添加变量来获取已检查顶点的数量，并在方法relax（e）中增加它。因此，如果我们假设变量名是：numberOfCheckedVertices

private int numberOfCheckedVertices;

这里应该增加变量的数量：

private void relax(DirectedEdge e) {
//increase the amount of visited vertices
numberOfCheckedVertices++;
int v = e.from(), w = e.to();
if (distTo[w] > distTo[v] + e.weight()) {
    distTo[w] = distTo[v] + e.weight();
    edgeTo[w] = e;
    if (pq.contains(w)) pq.decreaseKey(w, distTo[w]);
    else   {
        pq.insert(w, distTo[w]);                
    }            
}}

也许你会问那个方法里面的原因？因为当我们调用方法relax（e）时，你实际上已经检查了节点（顶点）

Dijkstra最短路径算法：查找已检查顶点的数量

2 个答案: