(源,目标)和它的类型(树,后,前进,交叉)?
你去吧。Java 中的代码
import java.util.ArrayList;
import java.util.List;
class Node {
public String name;
public List<Node> connections = new ArrayList<>();
boolean visited = false;
Node(String name) {
this.name = name;
}
}
class DFS {
// Main part.
public static void search(Node root) {
if (root == null) {
return;
}
root.visited = true;
for (Node node : root.connections) {
if (!node.visited) {
// Print result.
System.out.println(root.name + "->" + node.name);
search(node);
}
}
}
}
public class App {
public static void main(String[] args) {
Node a = new Node("a");
Node b = new Node("b");
Node c = new Node("c");
Node d = new Node("d");
Node e = new Node("e");
a.connections.add(b);
b.connections.add(a);
b.connections.add(c);
b.connections.add(d);
c.connections.add(b);
c.connections.add(d);
d.connections.add(b);
d.connections.add(c);
d.connections.add(e);
DFS.search(d);
}
}
好问题。
这是基于您作为评论发布的来源的解决方案。重要提示:开始/结束表上有一个错误,第三行第三列应该是"end[u] <end[v]"而不是"end[u]> end[v]"
void main(G, s){
for each node v in G{
explored[v]=false
parent[v]=null
start[v]=end[v]=null
}
Global clock = 0
DFS(G, s, s)
}
void DFS(G, s, parent){
explored[s] = true;
parent[s] = parent
start[s]=clock
clock++
for each u=(s,v){
if(explored[v] == false){
DFS(G, v)
}
print (s + "-->" + v +"type: " + getType(s,v))
}
end[s]=clock
clock++
}
String getType(s, v){
if(start[s]<start[v]){
if(end[s]>end[v]) return "Tree edge"
else return "Forward edge"
else{
if(end[s]<end[v]) return "Back edge"
else return "Cross edge"
}
}