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java - 获取2个节点之间的所有路径列表

转载 作者:行者123 更新时间:2023-12-01 13:14:07 27 4
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任何人都可以帮助我理解我在这里做错了什么。我想获取图中两个给定节点之间的所有路径的列表。任何帮助将非常感激。

public class City {
final private String id;

public City(String id) {
this.id = id;
}
public City getCity() {
return node;
}
public int getCost() {
return cost;
}
}

public class Route {

private final City node;
private final int cost;

public Route(City node, int cost) {
this.node = node;
this.cost = cost;
}
}

public class Graph {

private final List<City> cities;
private final List<Route> routes;
private final Map<City, List<Route>> myGraph = new HashMap<City,List<Route>>();
private Queue<City> visited = new LinkedList<City>();
private Queue<City> beenThere = new LinkedList<City>();

public Graph(List<City> vertexes, List<Route> edges) {
this.cities = vertexes;
this.routes = edges;
}

public String toString () {
StringBuffer s = new StringBuffer();
for (City c: myGraph.keySet())
System.out.println(c + " -> " + myGraph.get(c));
System.out.println();
return s.toString();
}

public void addNew (City vertex) {
if (myGraph.containsKey(vertex))
return;
myGraph.put(vertex, new ArrayList<Route>());
}

public void add (City from, City to, int cost) {
this.addNew(from);
this.addNew(to);
myGraph.get(from).add(new Route(to,cost));
myGraph.get(to).add(new Route(from,cost));
}

public boolean contains (City vertex) {
return myGraph.containsKey(vertex);
}

// public Route returnRoute

public Queue<City> getNeighbors(City city) {
Queue<City> neighbors = new LinkedList<City>();
for (City c : myGraph.keySet()) {
String c1 = c.toString();
String city1 = city.toString();
if (c1.equals(city1))
neighbors.add(myGraph.get(c).get(0).getCity());
}
return neighbors;
}

public Queue<City> checkRoute(City node) {
Queue<City> neighbors = new LinkedList<City>();
neighbors = this.getNeighbors(node);
for (City n : neighbors)
if(visited.contains(n))
break;
else
{
visited.add(n);
beenThere.add(n);
checkRoute(n);
}

return beenThere;
}

public List<Queue<City>> BFS(Graph graph, City from, City to) {
List<Queue<City>> rute = new ArrayList<Queue<City>>();
Queue<City> toVisit = new LinkedList<City>();
toVisit.add(from);
beenThere.add(from);
while(!toVisit.isEmpty()) {
City node = toVisit.remove();
visited.add(node);
Queue<City> neighbors = new LinkedList<City>();
neighbors = this.getNeighbors(node);
while(!neighbors.isEmpty()) {
visited.add(neighbors.element());
checkRoute(neighbors.remove());
}
if (beenThere.poll().equals(to))
rute.add(beenThere);
beenThere.clear();
beenThere.add(from);
//visited.clear();
}
return rute;
}

public void ReadFile() throws IOException {
String scan;
FileReader file = new FileReader("C:\\Users\\W7\\workspace\\AI-Lab1\\graph.txt");
BufferedReader br = new BufferedReader(file);
while((scan = br.readLine()) != null) {
String[] numberString =scan.split(" ");
String from = numberString[0];
City city1 = new City(from);
String to = numberString[1];
City city2 = new City(to);
int cost = Integer.parseInt(numberString[2]);
this.add(city1, city2, cost);
}
br.close();
}
}

城市是从文本文件 graph.txt 中读取的:

Bucuresti Atena 1700
Bucuresti Budapesta 800
Paris Bucuresti 2000
Madrid Berlin 500
Kiew Moscova 800
Oslo Stokolm 590
Bucuresti Berlin 1500
Budapesta Berlin 500

public class Start {

private static List<City> nodes;
private static List<Route> edges;

private static void run () throws IOException {
nodes = new ArrayList<City>();
edges = new ArrayList<Route>();
Graph graph = new Graph(nodes, edges);
graph.ReadFile();
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
while(true) {
System.out.println("Press:");
System.out.println("[1] VIEW THE GRAPH");
System.out.println("[2] GET THE SHORTEST PATH");
try {
String s = br.readLine();
if(s.equals("exit"))
return;
int option=Integer.parseInt(s);
if (option == 1 ) {
System.out.println("The graph is:" );
System.out.println(graph);
//getGraph();
System.out.println("========================");
}
if (option == 2) {
System.out.println("\tGive source vertex : ");
String source = br.readLine();
City from = new City(source);
System.out.println("\tGive destination vertex : ");
String destination = br.readLine();
City to = new City(destination);
//getShortestPath(graph,source,destination);
System.out.println(graph.BFS(graph, from, to));
System.out.println("========================");
}
if (option == 3) {
System.out.println("Please insert city:");
String c = br.readLine();
City city = new City(c);
System.out.println(graph.getNeighbors(city));
}
}catch(IOException e) {
e.printStackTrace();
}
}
}

public static void main(String[] args) throws IOException {
run();

}
}

这会返回一个空列表,我不明白为什么......

最佳答案

我注意到你有:

if (beenThere.poll().equals(to)) {
rute.add(beenThere);
}

要评估您自己的个人对象之一的相等性,您需要重写该类的 hashCode()equals() 方法。

否则,两个对象永远不会相等,除非它们是内存中完全相同的对象。将这两个方法添加到您的 City 类中:

@Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + ((id == null) ? 0 : id.hashCode());
return result;
}

@Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
City other = (City) obj;
if (id == null) {
if (other.id != null)
return false;
} else if (!id.equals(other.id))
return false;
return true;
}

另一个问题是,您正在清除 beenThere ,这也会清除 ArrayList 中的值,因为它们都引用内存中的同一对象。要理解这一点,请尝试使用以下代码,其中注释掉 clear() 方法以查看差异。

public static void main(String[] args) {
List<Queue<Integer>> rute = new ArrayList<Queue<Integer>>();
Queue<Integer> beenThere = new LinkedList<Integer>();

beenThere.add(1);
beenThere.add(2);
beenThere.add(3);
rute.add(beenThere);

beenThere.clear(); // run this code with this commented out to see the difference

System.out.println(rute.toString());
}

关于java - 获取2个节点之间的所有路径列表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/22597908/

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