问题
I am currently trying to traverse all paths from source to destination in a graph which uses adjacency matrix. I have been trying to do it in BFS way.Thanks for the help. I am getting only one path. How do I get to print other paths as well ?
public class AllPossiblePaths {
static int v;
static ArrayList<Integer> adj[];
public AllPossiblePaths(int v) {
this.v = v;
adj = new ArrayList[v];
for (int i = 0; i < v; i++) {
adj[i] = new ArrayList<>();
}
}
// add edge from u to v
public static void addEdge(int u, int v) {
adj[u].add(v);
}
public static void findpaths(int source, int destination) {
LinkedList<ArrayList<Integer>> q = new LinkedList<>();
boolean visited[] = new boolean[v];
LinkedList<Integer> queue = new LinkedList<Integer>();
queue.add(source);
visited[source] = true;
ArrayList<Integer> localPath = new ArrayList<>();
while (!queue.isEmpty()) {
// Dequeue a vertex from queue and print it
int src = queue.poll();
if (!localPath.contains(src)) {
localPath.add(src);
}
if (src == destination) {
System.out.println(localPath);
localPath.remove(localPath.size() - 1);
visited[src] = false;
}
Iterator<Integer> i = adj[src].listIterator();
while (i.hasNext()) {
int n = i.next();
if (!visited[n]) {
queue.add(n);
}
}
}
}
}
回答1:
Using the following class you can run a BFS to find a single path (findPath) or find multiple paths (findAllPaths). See comments:
import java.util.ArrayList;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.List;
public class AllPossiblePaths {
private boolean[] visited;
//keep track of nodes already included in a path
private boolean[] includedInPath;
private LinkedList<Integer> queue;
private int numberOfNodes;
private List<Integer>[] adj;
//to find a path you need to store the path that lead to it
private List<Integer>[] pathToNode;
public AllPossiblePaths(int numberOfNodes) {
this.numberOfNodes = numberOfNodes;
adj = new ArrayList[numberOfNodes];
pathToNode = new ArrayList[numberOfNodes];
for (int i = 0; i < numberOfNodes; i++) {
adj[i] = new ArrayList<>();
}
}
// add edge from u to v
public AllPossiblePaths addEdge(int from, int to) {
adj[from].add(to);
//unless unidirectional: //if a is connected to b
//than b should be connected to a
adj[to].add(from);
return this; //makes it convenient to add multiple edges
}
public void findPath(int source, int destination) {
System.out.println("------------Single path search---------------");
initializeSearch(source);
while (!queue.isEmpty()) {
// Dequeue a vertex from queue and print it
int src = queue.poll();
visited[src] = true;
if (src == destination) {
System.out.println("Path from "+source+" to "
+ destination+ " :- "+ pathToNode[src]);
break; //exit loop if target found
}
Iterator<Integer> i = adj[src].listIterator();
while (i.hasNext()) {
int n = i.next();
if (! visited[n] && ! queue.contains(n)) {
queue.add(n);
pathToNode[n].addAll(pathToNode[src]);
pathToNode[n].add(src);
}
}
}
}
public void findAllpaths(int source, int destination) {
System.out.println("-----------Multiple path search--------------");
includedInPath = new boolean[numberOfNodes];
initializeSearch(source);
int pathCounter = 0;
while(! allVisited() && !queue.isEmpty()) {
while (!queue.isEmpty()) {
// Dequeue a vertex from queue and print it
int src = queue.poll();
visited[src] = true;
if (src == destination) {
System.out.println("Path " + ++pathCounter + " from "+source+" to "
+ destination+ " :- "+ pathToNode[src]);
//mark nodes that are included in the path, so they will not be included
//in any other path
for(int i=1; i < pathToNode[src].size(); i++) {
includedInPath[pathToNode[src].get(i)] = true;
}
initializeSearch(source); //initialize before restarting
break; //exit loop if target found
}
Iterator<Integer> i = adj[src].listIterator();
while (i.hasNext()) {
int n = i.next();
if (! visited[n] && ! queue.contains(n)
&& ! includedInPath[n] /*ignore nodes already in a path*/) {
queue.add(n);
pathToNode[n].addAll(pathToNode[src]);
pathToNode[n].add(src);
}
}
}
}
}
private void initializeSearch(int source) {
queue = new LinkedList<>();
queue.add(source);
visited = new boolean[numberOfNodes];
for (int i = 0; i < numberOfNodes; i++) {
pathToNode[i]= new ArrayList<>();
}
}
private boolean allVisited() {
for( boolean b : visited) {
if(! b ) return false;
}
return true;
}
}
For testing it, consider this graph:
Run test:
public static void main(String[] args){
AllPossiblePaths app = new AllPossiblePaths(6);
app.addEdge(0, 4)
.addEdge(0, 1)
.addEdge(1, 2)
.addEdge(1, 4)
.addEdge(4, 3)
.addEdge(2, 3)
.addEdge(2, 5)
.addEdge(3, 5);
app.findPath(0,5);
app.findPath(5,0);
app.findAllpaths(0,5);
}
output:
回答2:
Apparently it is impossible to retrieve all paths from a given source to a given terminal via Breadth-First search. Consider the following class of graphs.
For any nonnegative integer n, let
V := {v_1,...,v2_n} // inner vertices
union
{s, t}, // source and terminal
E := { {v_i,v+2,} : i < 2n-2 } // horizontal edges
union
{ {v_i,v_i+3} : i < 2n-3, i is odd } // cross edges from top to bottom
union
{ {v_i,v_i+3} : i < 2n-3, i is even } // cross edges from bottom to top
union
{ {s,v_1}, {s,v_2}, {t,v_2n-1}, {t,v_2n} } // source and terminal
Informally, the graph consists out of two rows of vertices with n columns each, to the left there is a source node and to the right there is a terminal node. For each path from s to t, you can choose for each column to stay in the current row or to switch to the other row.
In total, there are 2^n different paths from s to t, as for each column there are two possibilities to chose the row.
On the other hand, Breadth-First search yields a runtime bound which is polynomial in the encoding length of the graph; this means that Breadth-first search, in general, cannot generate all possible paths from a given source to a given terminal. Furthermore, if the graph contains a cycle, the number of paths might be inifinite via repetition of the cycle.
来源：https://stackoverflow.com/questions/48699082/bfs-traversal-of-all-paths-in-graph-using-adjacency-list