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[英]Find Shortest Path in a directed weighted graph from list of sources and list of destinations
[英]How to represent a unweight directed graph and find the shortest path? Java
用Java表示图形的最佳方法是什么? 我这样做是这样的:
public class Node<T> {
public T data;
public LinkedList<Node> children;
public Node(T data) {
this.data = data;
this.children = new LinkedList<Node>(); //here there are the connected nodes of the node
}
public T getData() {
return this.data;
}
public void addArch(Node n) {
children.add(n);
}
public class Graph <T> {
private Node<T> source = new Node(null);
private Node<T> target = new Node(null);
private ArrayList<Node> nodes = new ArrayList<Node>();
public Graph() {}
public void addNode(T v) {
boolean visto = false;
for (Node n: nodes) {
if (n.getData().equals(v)) {
visto = true;
}
}
if (visto == false) {
nodes.add(new Node(v));
}
}
public void addEdge(T p, T a) throws NoSuchNodeException {
boolean visto = false;
boolean visto_secondo = false;
for (Node n: nodes) {
if (n.getData().equals(p)) {
visto = true;
}
}
for (Node n: nodes) {
if (n.getData().equals(a)) {
visto_secondo = true;
}
}
if (visto == false || visto_secondo == false) {
throw new NoSuchNodeException();
}
else {
for (Node n : nodes) {
if (p.equals(n.getData())) {
System.out.print(a);
n.addArch(new Node(a));
}
}
}
}
我必须找到最短的路径,但是好像没有添加拱门,为什么? 我也做了源和目标的设置和获取。 但是,我必须找到此源和目标之间的最短路径,使用什么算法? 我需要使用bfs来获得最短路径,但是我的问题是如何遍历拱门,我需要做一个递归函数
找到目标节点最短路径的最佳算法是迭代加深A * 。
如果启发式值是可接受的,它会找到到达目标节点的最短路径,这意味着它不会高估。
这是伪代码:
node current node
g the cost to reach current node
f estimated cost of the cheapest path (root..node..goal)
h(node) estimated cost of the cheapest path (node..goal)
cost(node, succ) step cost function
is_goal(node) goal test
successors(node) node expanding function, expand nodes ordered by g + h(node)
procedure ida_star(root)
bound := h(root)
loop
t := search(root, 0, bound)
if t = FOUND then return bound
if t = ∞ then return NOT_FOUND
bound := t
end loop
end procedure
function search(node, g, bound)
f := g + h(node)
if f > bound then return f
if is_goal(node) then return FOUND
min := ∞
for succ in successors(node) do
t := search(succ, g + cost(node, succ), bound)
if t = FOUND then return FOUND
if t < min then min := t
end for
return min
end function
g表示达到当前状态的移动次数,h表示达到目标状态的估计移动次数。 f := g + h(node)
启发式值越接近实际移动次数,算法就越快
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