据我所理解，
但是，通过使用
由于我理解
那么，为什么
我希望这个问题可以在不发布代码的情况下得到回答，但是如果需要，我当然可以发布完整的程序。
该程序使用深度优先搜索从给定的起点到给定的终点搜索迷宫。解决方案是存在的。主逻辑在
我引用的数字（“30,000个任务，平均700个线程”）来自搜索5000x5000个单元的迷宫。因此，以下是关键代码：
ForkJoinPool池会创建一定数量的线程（默认为核心数），并且不会再创建更多的线程（除非应用程序通过使用managedBlock来指示需要更多线程）。但是，通过使用
ForkJoinPool.getPoolSize()，我发现在一个创建了30,000个任务（RecursiveAction）的程序中，执行这些任务的ForkJoinPool平均使用700个线程（每次创建任务时都计算线程数）。这些任务不涉及I/O，只涉及纯计算；任务之间唯一的同步是调用ForkJoinTask.join()和访问AtomicBoolean，即没有线程阻塞操作。由于我理解
join()不会阻塞调用线程，所以池中任何线程都不应该被阻塞，因此（我原以为）不需要创建更多的线程（尽管仍然发生了这种情况）。那么，为什么
ForkJoinPool会创建这么多线程？哪些因素决定了创建线程的数量？我希望这个问题可以在不发布代码的情况下得到回答，但是如果需要，我当然可以发布完整的程序。
该程序使用深度优先搜索从给定的起点到给定的终点搜索迷宫。解决方案是存在的。主逻辑在
SolverTask的compute()方法中：一个RecursiveAction从某个给定点开始，并继续使用当前点可达的所有相邻点。它不会在每个分支点创建新的SolverTask（这将创建太多任务），而是将除一个以外的所有相邻点压入回溯堆栈以后处理，并仅使用未推送到堆栈的一个邻居点继续执行。一旦它通过这种方式到达死路，就会弹出最近推送到回溯堆栈中的点，并从那里继续搜索（相应地缩短从任务的起点建立的路径）。一旦任务发现其回溯堆栈大于某个特定阈值，它将创建一个新任务；从那时起，该任务将继续从其回溯堆栈中弹出，直到完全清空堆栈。当到达分支点时，它不会再将任何其他点压入其堆栈，而是为每个此类点创建一个新的任务。因此，可以使用堆栈限制阈值调整任务大小。我引用的数字（“30,000个任务，平均700个线程”）来自搜索5000x5000个单元的迷宫。因此，以下是关键代码：
class SolverTask extends RecursiveTask<ArrayDeque<Point>> {
// Once the backtrack stack has reached this size, the current task
// will never add another cell to it, but create a new task for each
// newly discovered branch:
private static final int MAX_BACKTRACK_CELLS = 100*1000;
/**
* @return Tries to compute a path through the maze from local start to end
* and returns that (or null if no such path found)
*/
@Override
public ArrayDeque<Point> compute() {
// Is this task still accepting new branches for processing on its own,
// or will it create new tasks to handle those?
boolean stillAcceptingNewBranches = true;
Point current = localStart;
ArrayDeque<Point> pathFromLocalStart = new ArrayDeque<Point>(); // Path from localStart to (including) current
ArrayDeque<PointAndDirection> backtrackStack = new ArrayDeque<PointAndDirection>();
// Used as a stack: Branches not yet taken; solver will backtrack to these branching points later
Direction[] allDirections = Direction.values();
while (!current.equals(end)) {
pathFromLocalStart.addLast(current);
// Collect current's unvisited neighbors in random order:
ArrayDeque<PointAndDirection> neighborsToVisit = new ArrayDeque<PointAndDirection>(allDirections.length);
for (Direction directionToNeighbor: allDirections) {
Point neighbor = current.getNeighbor(directionToNeighbor);
// contains() and hasPassage() are read-only methods and thus need no synchronization
if (maze.contains(neighbor) && maze.hasPassage(current, neighbor) && maze.visit(neighbor))
neighborsToVisit.add(new PointAndDirection(neighbor, directionToNeighbor.opposite));
}
// Process unvisited neighbors
if (neighborsToVisit.size() == 1) {
// Current node is no branch: Continue with that neighbor
current = neighborsToVisit.getFirst().getPoint();
continue;
}
if (neighborsToVisit.size() >= 2) {
// Current node is a branch
if (stillAcceptingNewBranches) {
current = neighborsToVisit.removeLast().getPoint();
// Push all neighbors except one on the backtrack stack for later processing
for(PointAndDirection neighborAndDirection: neighborsToVisit)
backtrackStack.push(neighborAndDirection);
if (backtrackStack.size() > MAX_BACKTRACK_CELLS)
stillAcceptingNewBranches = false;
// Continue with the one neighbor that was not pushed onto the backtrack stack
continue;
} else {
// Current node is a branch point, but this task does not accept new branches any more:
// Create new task for each neighbor to visit and wait for the end of those tasks
SolverTask[] subTasks = new SolverTask[neighborsToVisit.size()];
int t = 0;
for(PointAndDirection neighborAndDirection: neighborsToVisit) {
SolverTask task = new SolverTask(neighborAndDirection.getPoint(), end, maze);
task.fork();
subTasks[t++] = task;
}
for (SolverTask task: subTasks) {
ArrayDeque<Point> subTaskResult = null;
try {
subTaskResult = task.join();
} catch (CancellationException e) {
// Nothing to do here: Another task has found the solution and cancelled all other tasks
}
catch (Exception e) {
e.printStackTrace();
}
if (subTaskResult != null) { // subtask found solution
pathFromLocalStart.addAll(subTaskResult);
// No need to wait for the other subtasks once a solution has been found
return pathFromLocalStart;
}
} // for subTasks
} // else (not accepting any more branches)
} // if (current node is a branch)
// Current node is dead end or all its neighbors lead to dead ends:
// Continue with a node from the backtracking stack, if any is left:
if (backtrackStack.isEmpty()) {
return null; // No more backtracking avaible: No solution exists => end of this task
}
// Backtrack: Continue with cell saved at latest branching point:
PointAndDirection pd = backtrackStack.pop();
current = pd.getPoint();
Point branchingPoint = current.getNeighbor(pd.getDirectionToBranchingPoint());
// DEBUG System.out.println("Backtracking to " + branchingPoint);
// Remove the dead end from the top of pathSoFar, i.e. all cells after branchingPoint:
while (!pathFromLocalStart.peekLast().equals(branchingPoint)) {
// DEBUG System.out.println(" Going back before " + pathSoFar.peekLast());
pathFromLocalStart.removeLast();
}
// continue while loop with newly popped current
} // while (current ...
if (!current.equals(end)) {
// this task was interrupted by another one that already found the solution
// and should end now therefore:
return null;
} else {
// Found the solution path:
pathFromLocalStart.addLast(current);
return pathFromLocalStart;
}
} // compute()
} // class SolverTask
@SuppressWarnings("serial")
public class ParallelMaze {
// for each cell in the maze: Has the solver visited it yet?
private final AtomicBoolean[][] visited;
/**
* Atomically marks this point as visited unless visited before
* @return whether the point was visited for the first time, i.e. whether it could be marked
*/
boolean visit(Point p) {
return visited[p.getX()][p.getY()].compareAndSet(false, true);
}
public static void main(String[] args) {
ForkJoinPool pool = new ForkJoinPool();
ParallelMaze maze = new ParallelMaze(width, height, new Point(width-1, 0), new Point(0, height-1));
// Start initial task
long startTime = System.currentTimeMillis();
// since SolverTask.compute() expects its starting point already visited,
// must do that explicitly for the global starting point:
maze.visit(maze.start);
maze.solution = pool.invoke(new SolverTask(maze.start, maze.end, maze));
// One solution is enough: Stop all tasks that are still running
pool.shutdownNow();
pool.awaitTermination(Integer.MAX_VALUE, TimeUnit.DAYS);
long endTime = System.currentTimeMillis();
System.out.println("Computed solution of length " + maze.solution.size() + " to maze of size " +
width + "x" + height + " in " + ((float)(endTime - startTime))/1000 + "s.");
}