早上好!用DFS查找图中的所有路径
我正在开发一种算法来查找无向图中的所有路径。我目前正在使用带回溯的DFS algortihm来尝试这样做。这是我现在的代码:
import java.util.*;
public class dfs {
private static Map<Integer, LinkedHashSet<Integer>> map = new HashMap<Integer, LinkedHashSet<Integer>>();
private int startNode;
private int numLinks;
public dfs(int startNode, int numLinks) {
super();
this.startNode = startNode;
this.numLinks = numLinks;
}
public void addEdge(int source, int destiny) {
LinkedHashSet<Integer> adjacente = map.get(source);
if(adjacente==null) {
adjacente = new LinkedHashSet<Integer>();
map.put(source, adjacente);
}
adjacente.add(destiny);
}
public void addLink(int source, int destiny) {
addEdge(source, destiny);
addEdge(destiny, source);
}
public LinkedList<Integer> adjacentNodes(int last) {
LinkedHashSet<Integer> adjacente = map.get(last);
System.out.println("adjacentes:" + adjacente);
if(adjacente==null) {
return new LinkedList<Integer>();
}
return new LinkedList<Integer>(adjacente);
}
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int numVertices = input.nextInt();
int numLinks = input.nextInt();
int startNode = input.nextInt();
int endNode = startNode;
dfs mapa = new dfs(startNode, numLinks);
for(int i = 0; i<numLinks; i++){
mapa.addLink(input.nextInt(), input.nextInt());
}
List<ArrayList<Integer>> paths = new ArrayList<ArrayList<Integer>>();
List<Integer> visited = new ArrayList<Integer>();
visited.add(startNode);
Integer currentNode = 0;
Iterator it = map.entrySet().iterator();
while (it.hasNext()) {
Map.Entry pairs = (Map.Entry)it.next();
currentNode = (Integer) pairs.getKey();
//System.out.println("Current Node:" + currentNode);
mapa.findAllPaths(mapa, visited, paths, currentNode);
}
}
private void findAllPaths(dfs mapa, List<Integer> visited,
List<ArrayList<Integer>> paths, Integer currentNode) {
if (currentNode.equals(startNode)) {
paths.add(new ArrayList<Integer>(visited));
LinkedList<Integer> nodes = mapa.adjacentNodes(currentNode);
//System.out.println("visited:" + visited);
for (Integer node : nodes) {
//System.out.println("nodes:" + nodes);
List<Integer> temp = new ArrayList<Integer>();
temp.addAll(visited);
temp.add(node);
findAllPaths(mapa, temp, paths, node);
}
}
else {
LinkedList<Integer> nodes = mapa.adjacentNodes(currentNode);
System.out.println("currentNode:" + currentNode);
//System.out.println("nodes:" + nodes);
for (Integer node : nodes) {
if (visited.contains(node)) {
continue;
}
List<Integer> temp = new ArrayList<Integer>();
temp.addAll(visited);
System.out.println("visited:" + visited);
temp.add(node);
findAllPaths(mapa, temp, paths, node);
}
}
}
}
该程序在他的输入接收整数。第一个是节点数量,第二个是链接数量,第三个是开始节点和结束音符,它们是相同的。所有后面的整数表示节点之间的连接。
问题是,该算法找到的所有路径只访问一次单个节点。我想要的是查找所有访问每个连接的路径的算法。 关于如何做到这一点的任何想法?
节点之间的连接如何表示为整数?我不太明白。 – 2012-03-21 11:03:32
一个例子,程序收到“1 2 3 3 4 5 6”1是节点的数量,2是链接的数量,3是起始节点,3 4是从节点3到节点4的连接,5 6是从节点5到6的连接。 – 2012-03-21 11:42:13