具有信号量的Java线程同步

Question

我在完成家庭作业时遇到了麻烦，因此在经过数小时的尝试和谷歌搜索后，我除了寻求其他帮助外别无选择。我有这段代码（不是整个程序，也有一些类和数据包以图形方式模拟了此代码的工作方式，但我认为不必在此处发布它）：

public class KidsAndTramboline {

 private Tramboline tramboline = new Tramboline(300, 5);

 private class Tramboline {

  private Semaphore[] mutexes;
  private int[] counters;
  private Semaphore free;
  private Semaphore weight;
  private Semaphore kids;

  public Tramboline(int maxWeight, int maxKids) {
   mutexes = new Semaphore[] {
     new Semaphore(1),
     new Semaphore(1)
   };
   counters = new int[] {
     0,
     0
   };
   free = new Semaphore(1);
   weight = new Semaphore(maxWeight);
   kids = new Semaphore(maxKids);
  }

  public void getOn(Kid kid) {

   int sex = kid.sex == Sex.MALE ? 1 : 0;

   mutexes[sex].acquireUninterruptibly();
   counters[sex]++;
   if (counters[sex] == 1) {
    free.acquireUninterruptibly();
   }
   mutexes[sex].release();

   weight.acquireUninterruptibly(kid.weight);

   kids.acquireUninterruptibly(1);

  }

  public void getOff(Kid kid) {

   int sex = kid.sex == Sex.MALE ? 1 : 0;

   mutexes[sex].acquireUninterruptibly();
   counters[sex]--;
   if (counters[sex] == 0) {
    free.release();
   }
   mutexes[sex].release();

   weight.release(kid.weight);

   kids.release(1);;

  }
 }

 private class Kid extends UtilThread {

     private final Sex sex = getSex();
     private final int weight = getWeight();

        @Override
        protected void step() {
            resting();
         tramboline.getOn(this);
            try {
                jumping();
            } finally {
    tramboline.getOff(this);
   }
        }
    }

 private enum Sex {
  MALE, FEMALE;
 }
}

与孩子和蹦床有关，这是某种读者和作家的问题。关键是男孩在跳，女孩必须等待，反之亦然。我要做的是，当男孩和女孩在一起等待时（蹦床空无一人），应优先考虑女孩，以便他们在男孩之前上车。你能帮我解决这个问题吗？谢谢。