迭代拓扑搜索 (DFS)

Question

如何在有向无环图上完成迭代 dfs 拓扑排序？

这是一个顶点

class Vertex {
  List<Vertex> adj = new ArrayList<>();
  char val;

  Vertex(char val) {this.val = val;}
}

递归解决方案很简单，使用集合来标记访问的节点和堆栈来对顶点进行排序：

List<Vertex> sortRecursive(List<Vertex> vertices) {
  Deque<Vertex> stack = new ArrayDeque<>();
  Set<Vertex> visited = new HashSet<>();
  for (Vertex Vertex : vertices) {
    if (visited.contains(Vertex)) continue;
    sortRecursiveHelper(stack, visited, Vertex);
  }
  List<Vertex> output = new ArrayList<>();
  while (!stack.isEmpty()) output.add(stack.removeFirst());
  return output;
}

void sortRecursiveHelper(Deque<Vertex> stack, Set<Vertex> visited, Vertex vertex) {
  visited.add(vertex);
  for (Vertex vv : vertex.adj) {
    if (visited.contains(vv)) continue;
    sortRecursiveHelper(stack, visited, vv);
  }
  stack.addFirst(vertex);
}

这是司机：

Vertex a = new Vertex('A');
Vertex b = new Vertex('B');
Vertex c = new Vertex('C');
Vertex d = new Vertex('D');
Vertex e = new Vertex('E');
Vertex f = new Vertex('F');
Vertex g = new Vertex('G');


a.adj.add(c);
b.adj.add(c);
b.adj.add(e);
c.adj.add(d);
d.adj.add(f);
e.adj.add(f);
f.adj.add(g);

List<Vertex> output = sortRecursive(Arrays.asList(d, a, e, g, f, b, c));
System.out.println(output);

Answer 1

您可以保留一堆活动顶点和尚未处理的第一个子节点的索引来模拟递归：

while stack.non_empty()
    if stack.top().second == graph[stack.top().first].size:
        // We pop the vertex here, so we add it to the answer list
        sorted_order.add(stack.top().first)
        stack.pop_back()
    else:
        // We get the next child and move increase the index
        // so that it points to the unprocessed child
        next_child = graph[stack.top().first][stack().top().second]
        stack.top().second += 1
        // If we need to go to the child, we push it to the
        // stack instead of making a recursive call
        if not next_child is visited:
            mark next_child as visited
            stack.push(pair(next_child, 0))