本文介绍了统一成本搜索实施的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
在Udacity中观看AI简介课程后，我正在尝试实施统一成本搜索。但是，我的算法没有得到正确的路径。在发布之前一直在尝试一整天。我添加了一个地图来帮助可视化场景。该算法应找到从Arad到Bucharest的最短加权路径
I am trying to implement the Uniform Cost Search after watching the "Intro to AI" course in Udacity. However, my algorithm is not getting the correct path. Have been trying the whole day before posting here. I have added a map to help to visualize the scene. The algorithm should find the shortest weighted path from Arad to Bucharest
import java.util.PriorityQueue;
import java.util.HashSet;
import java.util.Set;
import java.util.Collections;
import java.util.List;
import java.util.ArrayList;
import java.util.Comparator;
//diff between uniform cost search and dijkstra algo is that UCS has a goal
public class UniformCostSearchAlgo{
public static void main(String[] args){
//initialize the graph base on the Romania map
Node n1 = new Node("Arad");
Node n2 = new Node("Zerind");
Node n3 = new Node("Oradea");
Node n4 = new Node("Sibiu");
Node n5 = new Node("Fagaras");
Node n6 = new Node("Rimnicu Vilcea");
Node n7 = new Node("Pitesti");
Node n8 = new Node("Timisoara");
Node n9 = new Node("Lugoj");
Node n10 = new Node("Mehadia");
Node n11 = new Node("Drobeta");
Node n12 = new Node("Craiova");
Node n13 = new Node("Bucharest");
Node n14 = new Node("Giurgiu");
//initialize the edges
n1.adjacencies = new Edge[]{
new Edge(n2,75),
new Edge(n4,140),
new Edge(n8,118)
};
n2.adjacencies = new Edge[]{
new Edge(n1,75),
new Edge(n3,71)
};
n3.adjacencies = new Edge[]{
new Edge(n2,71),
new Edge(n4,151)
};
n4.adjacencies = new Edge[]{
new Edge(n1,140),
new Edge(n5,99),
new Edge(n3,151),
new Edge(n6,80),
};
n5.adjacencies = new Edge[]{
new Edge(n4,99),
new Edge(n13,211)
};
n6.adjacencies = new Edge[]{
new Edge(n4,80),
new Edge(n7,97),
new Edge(n12,146)
};
n7.adjacencies = new Edge[]{
new Edge(n6,97),
new Edge(n13,101),
new Edge(n12,138)
};
n8.adjacencies = new Edge[]{
new Edge(n1,118),
new Edge(n9,111)
};
n9.adjacencies = new Edge[]{
new Edge(n8,111),
new Edge(n10,70)
};
n10.adjacencies = new Edge[]{
new Edge(n9,70),
new Edge(n11,75)
};
n11.adjacencies = new Edge[]{
new Edge(n10,75),
new Edge(n12,120)
};
n12.adjacencies = new Edge[]{
new Edge(n11,120),
new Edge(n6,146),
new Edge(n7,138)
};
n13.adjacencies = new Edge[]{
new Edge(n7,101),
new Edge(n14,90),
new Edge(n5,211)
};
n14.adjacencies = new Edge[]{
new Edge(n13,90)
};
UniformCostSearch(n1,n13);
List<Node> path = printPath(n13);
System.out.println("Path: " + path);
}
public static void UniformCostSearch(Node source, Node goal){
source.pathCost = 0;
PriorityQueue<Node> queue = new PriorityQueue<Node>(20,
new Comparator<Node>(){
//override compare method
public int compare(Node i, Node j){
if(i.pathCost > j.pathCost){
return 1;
}
else if (i.pathCost < j.pathCost){
return -1;
}
else{
return 0;
}
}
}
);
queue.add(source);
Set<Node> explored = new HashSet<Node>();
boolean found = false;
//while frontier is not empty
do{
Node current = queue.poll();
explored.add(current);
if(current.value.equals(goal.value)){
found = true;
}
for(Edge e: current.adjacencies){
Node child = e.target;
double cost = e.cost;
child.pathCost = current.pathCost + cost;
if(!explored.contains(child) && !queue.contains(child)){
child.parent = current;
queue.add(child);
System.out.println(child);
System.out.println(queue);
System.out.println();
}
else if((queue.contains(child))&&(child.pathCost>current.pathCost)){
child.parent=current;
current = child;
}
}
}while(!queue.isEmpty());
}
public static List<Node> printPath(Node target){
List<Node> path = new ArrayList<Node>();
for(Node node = target; node!=null; node = node.parent){
path.add(node);
}
Collections.reverse(path);
return path;
}
}
class Node{
public final String value;
public double pathCost;
public Edge[] adjacencies;
public Node parent;
public Node(String val){
value = val;
}
public String toString(){
return value;
}
}
class Edge{
public final double cost;
public final Node target;
public Edge(Node targetNode, double costVal){
cost = costVal;
target = targetNode;
}
}
推荐答案
更改此部分代码
else if((queue.contains(child))&&(child.pathCost>current.pathCost)){
child.parent=current;
current = child;
}
到
else if((queue.contains(child))&&(child.pathCost>current.pathCost+cost)){
child.parent=current;
child.pathCost = current.pathCost+cost;
queue.remove(child);
queue.add(child);
}
这篇关于统一成本搜索实施的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！