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。
我想做的是计算一条路径的权重,这条路径的长度和已检查的顶点数。 我已成功找到前两个,但找到已检查的顶点数量并没有真正起作用。
举例说明我想做什么:
我想找出下图中有多少个方格被红线选中: http://prntscr.com/9921cv
我们非常欢迎任何有关如何开始的线索。
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 ≤ <tt>s</tt> ≤ <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;
}
}
最佳答案
finding the amount of vertices that have been checked isn't really working.
每次从队列中弹出一个顶点时,增加一个计数器(我们称它为 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,你应该改进所有变量和方法的名称。很难阅读您的代码。
关于java - Dijkstra最短路径算法: finding number of checked vertices,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/34020877/