How to use

Graph

in

org.antlr.misc

 public boolean isPath(Graph graph, Node start, Node target) {
  for (Node node : graph.getNode()) {
    if (node != null) {
      node.state = State.Unvisited;
    }
  }

  dfs(start);

  return target.state == State.Visited;
}

  public void DFS(Node<T> n, Set<Node<T>> visited, ArrayList<T> sorted) {
    if ( visited.contains(n) ) return;
    visited.add(n);
    if ( n.edges!=null ) {
      for (Node<T> target : n.edges) {
        DFS(target, visited, sorted);
      }
    }
    sorted.add(n.payload);
  }
}

public void sortGrammarFiles() throws IOException {
  Graph<String> g = new Graph<String>();
  List<String> missingFiles = new ArrayList<String>();
  for (String gfile : grammarFileNames) {
      String grammarName = grammar.getGrammarName();
      if ( vocabName!=null ) g.addEdge(gfile, vocabName+CodeGenerator.VOCAB_FILE_EXTENSION);
      g.addEdge(grammarName+CodeGenerator.VOCAB_FILE_EXTENSION, gfile);
  List<String> sorted = g.sort();

public void sortGrammarFiles() throws IOException {
  Graph g = new Graph();
  List<String> missingFiles = new ArrayList<String>();
  for (String gfile : grammarFileNames) {
      String grammarName = grammar.getGrammarName();
      if ( vocabName!=null ) g.addEdge(gfile, vocabName+CodeGenerator.VOCAB_FILE_EXTENSION);
      g.addEdge(grammarName+CodeGenerator.VOCAB_FILE_EXTENSION, gfile);
  List<Object> sorted = g.sort();

public void sortGrammarFiles() throws IOException {
  Graph g = new Graph();
  List<String> missingFiles = new ArrayList<String>();
  for (String gfile : grammarFileNames) {
      String grammarName = grammar.getGrammarName();
      if ( vocabName!=null ) g.addEdge(gfile, vocabName+CodeGenerator.VOCAB_FILE_EXTENSION);
      g.addEdge(grammarName+CodeGenerator.VOCAB_FILE_EXTENSION, gfile);
  List<Object> sorted = g.sort();

  public void DFS(Node<T> n, Set<Node<T>> visited, ArrayList<T> sorted) {
    if ( visited.contains(n) ) return;
    visited.add(n);
    if ( n.edges!=null ) {
      for (Node<T> target : n.edges) {
        DFS(target, visited, sorted);
      }
    }
    sorted.add(n.payload);
  }
}

public void addEdge(T a, T b) {
  //System.out.println("add edge "+a+" to "+b);
  Node<T> a_node = getNode(a);
  Node<T> b_node = getNode(b);
  a_node.addEdge(b_node);
}

public void sortGrammarFiles() throws IOException {
  Graph<String> g = new Graph<String>();
  List<String> missingFiles = new ArrayList<String>();
  for (String gfile : grammarFileNames) {
      String grammarName = grammar.getGrammarName();
      if ( vocabName!=null ) g.addEdge(gfile, vocabName+CodeGenerator.VOCAB_FILE_EXTENSION);
      g.addEdge(grammarName+CodeGenerator.VOCAB_FILE_EXTENSION, gfile);
  List<String> sorted = g.sort();

  public void DFS(Node n, Set<Node> visited, ArrayList<Object> sorted) {
    if ( visited.contains(n) ) return;
    visited.add(n);
    if ( n.edges!=null ) {
      for (Iterator it = n.edges.iterator(); it.hasNext();) {
        Node target = (Node) it.next();
        DFS(target, visited, sorted);
      }
    }
    sorted.add(n.payload);
  }
}

public void addEdge(Object a, Object b) {
  //System.out.println("add edge "+a+" to "+b);
  Node a_node = getNode(a);
  Node b_node = getNode(b);
  a_node.addEdge(b_node);
}

  public void DFS(Node n, Set<Node> visited, ArrayList<Object> sorted) {
    if ( visited.contains(n) ) return;
    visited.add(n);
    if ( n.edges!=null ) {
      for (Iterator it = n.edges.iterator(); it.hasNext();) {
        Node target = (Node) it.next();
        DFS(target, visited, sorted);
      }
    }
    sorted.add(n.payload);
  }
}

public void addEdge(T a, T b) {
  //System.out.println("add edge "+a+" to "+b);
  Node<T> a_node = getNode(a);
  Node<T> b_node = getNode(b);
  a_node.addEdge(b_node);
}

/** DFS-based topological sort.  A valid sort is the reverse of
 *  the post-order DFA traversal.  Amazingly simple but true.
 *  For sorting, I'm not following convention here since ANTLR
 *  needs the opposite.  Here's what I assume for sorting:
 *
 *    If there exists an edge u -> v then u depends on v and v
 *    must happen before u.
 *
 *  So if this gives nonreversed postorder traversal, I get the order
 *  I want.
 */
public List<Object> sort() {
  Set<Node> visited = new HashSet<Node>();
  ArrayList<Object> sorted = new ArrayList<Object>();
  while ( visited.size() < nodes.size() ) {
    // pick any unvisited node, n
    Node n = null;
    for (Iterator it = nodes.values().iterator(); it.hasNext();) {
      n = (Node)it.next();
      if ( !visited.contains(n) ) break;
    }
    DFS(n, visited, sorted);
  }
  return sorted;
}

public void addEdge(Object a, Object b) {
  //System.out.println("add edge "+a+" to "+b);
  Node a_node = getNode(a);
  Node b_node = getNode(b);
  a_node.addEdge(b_node);
}

/** DFS-based topological sort.  A valid sort is the reverse of
 *  the post-order DFA traversal.  Amazingly simple but true.
 *  For sorting, I'm not following convention here since ANTLR
 *  needs the opposite.  Here's what I assume for sorting:
 *
 *    If there exists an edge u → v then u depends on v and v
 *    must happen before u.
 *
 *  So if this gives nonreversed postorder traversal, I get the order
 *  I want.
 */
public List<T> sort() {
  Set<Node<T>> visited = new OrderedHashSet<Node<T>>();
  ArrayList<T> sorted = new ArrayList<T>();
  while ( visited.size() < nodes.size() ) {
    // pick any unvisited node, n
    Node<T> n = null;
    for (Node<T> tNode : nodes.values()) {
      n = tNode;
      if ( !visited.contains(n) ) break;
    }
    DFS(n, visited, sorted);
  }
  return sorted;
}

/** DFS-based topological sort.  A valid sort is the reverse of
 *  the post-order DFA traversal.  Amazingly simple but true.
 *  For sorting, I'm not following convention here since ANTLR
 *  needs the opposite.  Here's what I assume for sorting:
 *
 *    If there exists an edge u -> v then u depends on v and v
 *    must happen before u.
 *
 *  So if this gives nonreversed postorder traversal, I get the order
 *  I want.
 */
public List<Object> sort() {
  Set<Node> visited = new OrderedHashSet<Node>();
  ArrayList<Object> sorted = new ArrayList<Object>();
  while ( visited.size() < nodes.size() ) {
    // pick any unvisited node, n
    Node n = null;
    for (Iterator it = nodes.values().iterator(); it.hasNext();) {
      n = (Node)it.next();
      if ( !visited.contains(n) ) break;
    }
    DFS(n, visited, sorted);
  }
  return sorted;
}

/** DFS-based topological sort.  A valid sort is the reverse of
 *  the post-order DFA traversal.  Amazingly simple but true.
 *  For sorting, I'm not following convention here since ANTLR
 *  needs the opposite.  Here's what I assume for sorting:
 *
 *    If there exists an edge u → v then u depends on v and v
 *    must happen before u.
 *
 *  So if this gives nonreversed postorder traversal, I get the order
 *  I want.
 */
public List<T> sort() {
  Set<Node<T>> visited = new OrderedHashSet<Node<T>>();
  ArrayList<T> sorted = new ArrayList<T>();
  while ( visited.size() < nodes.size() ) {
    // pick any unvisited node, n
    Node<T> n = null;
    for (Node<T> tNode : nodes.values()) {
      n = tNode;
      if ( !visited.contains(n) ) break;
    }
    DFS(n, visited, sorted);
  }
  return sorted;
}