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问猜测同时运行的线程的下一个位置
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Stack Overflow用户

提问于 2011-12-11 04:11:55

回答 5查看 183关注 0票数 1

我正在做一个迷宫，两个线程正在处理simultaneously.The问题，它们不能在同一时间共享相同的位置。我不知道我该怎么做。是否有可能知道它们下一步将移动到哪里，并防止其他线程移动到该位置？请给我一个想法。谢谢。

代码如下：

        public int[][] visitedCell=new int[15][15]; // holds how many times visited for each cell
    public boolean cntrl=true,repeat=true,end;
    public int r=0;
    public int cnt;
    public boolean find;
    public void  decision(int posX,int posY){
       int distanceToExit;
       cnt=0; // holds how many cell is free around the cell which thread is on
       r=0;
       end=false; // checks robot found exit
       find=false; // checks suitable cell found to move in next step
       posX=posX/40; // all cells are 40*40 dimension.since it is divided 40 to find poisition
       posY=posY/40; // found y position
       int[][] neighbours={{posX,posY+1},{posX+1,posY+1},{posX+1,posY},{posX+1,posY-1},{posX,posY-1},{posX-1,posY-1},{posX-1,posY},{posX-1,posY+1}}; // all 8 neighbours of a cell
       int[][] freeCellChoises = new int[8][2]; // holds free cells to move
       int[][] distanceCell=new int[8][2];
       for(int i=0;i<8;i++){ // checks which neighbour cells are free
             if((neighbours[i][0] >0 && neighbours[i][0] <14) && (neighbours[i][1] >0 && neighbours[i][1] < 14) || (neighbours[i][0]==1 && neighbours[i][1]==14) || (neighbours[i][0]==14 && neighbours[i][1]==1) ) // [1,14] = enter position and [14,1]= exit position
             {
               if(Draw.paintArray[neighbours[i][0]][neighbours[i][1]]==0){// cell is free.it is eligible
                   freeCellChoises[cnt][0]=neighbours[i][0]; // load eligible cells this array
                   freeCellChoises[cnt][1]=neighbours[i][1];
                   distanceToExit=(int) Math.sqrt((Math.pow(neighbours[i][0]-560, 2)+Math.pow(neighbours[i][1]-40,2)));
                   distanceCell[cnt][0]=cnt;
                   distanceCell[cnt][1]=distanceToExit;
                   cnt++;}
             }
       } // eligible cells are ready anymore
       if(Frame.radButSel==1){ // random movement
           int no=rndm.nextInt(cnt); // choose one of the eligible cell randomly
           x=freeCellChoises[no][0] * 40;
           y=freeCellChoises[no][1] * 40;
       }
       if(Frame.radButSel==2){ // closed way movement ( find the most clodes cell to the exit ) .Exit is [14,1].So x must be max, y must be min to a cell has priority
           int maxX=freeCellChoises[0][0];
           int minY=freeCellChoises[0][1];
           int selection1=0,selection2=0;
           for(int i=0;i<cnt;i++){ // x i byk y si kck  sec
               if(freeCellChoises[i][0]> maxX){
                   maxX=freeCellChoises[i][0];
                   selection1=i;}
               if(freeCellChoises[i][1]<minY){
                   minY=freeCellChoises[i][1];
                   selection2=i;
               }


           }
         if(cnt!=0) // checks there is a priority cell
             r=rndm.nextInt(2)+1; // selects one of the priority cell

         if(r==1 && visitedCell[freeCellChoises[selection1][0]][freeCellChoises[selection1][1]] <2){ //selection1.same cell musnt be visited more than 2 times
               x=freeCellChoises[selection1][0] * 40;
               y=freeCellChoises[selection1][1] * 40;}

         else if(r==2 && visitedCell[freeCellChoises[selection2][0]][freeCellChoises[selection2][1]] <2){//selection2
              x=freeCellChoises[selection2][0] * 40;
              y=freeCellChoises[selection2][1] * 40;}

          else{ // applies when there is not any priority cell
               System.out.println("oncelik yok");
               int repeat =0;
               while(repeat<cnt){
                  r=rndm.nextInt(cnt); // choose one of the eligible cell
                  x=freeCellChoises[r][0] * 40;
                  y=freeCellChoises[r][1] * 40;
                  if(visitedCell[freeCellChoises[r][0]][freeCellChoises[r][1]] <2){
                      repeat=10;
                  }
                  else
                      repeat++;

               }System.out.println("x="+x+"y="+y);

          }
      }
      if(Frame.radButSel==3){

      }

      if(x==560 && y==40){ // checks decided cell is exit point
           Action.pool.shutdownNow();// thread finished
           end=true;
           Main.butAct++; // when butAct=2 , "RESULT" button will be active
           timer.stopTime();} // stops time for the thread

      distance=(int) Math.sqrt(Math.pow(x-560,2) + Math.pow(y-40, 2));// calculates distance between thread - exit

}

public Action() throws InterruptedException{
    pool=Executors.newFixedThreadPool(2); // two thread in the pool
    robot1=new Robot(40,560); // starts enter position
    robot2=new Robot(40,560); // starts enter position
    pool.submit(robot1); // loads robot1 to pool
    pool.submit(robot2);// loadss robot2 to pool
}
public void run() {
    while(true){ // run threads always
    try {
        Frame.worker.pauseIfNeeded();} // checks whether pause button is pressed
    catch (InterruptedException ex) {
        Logger.getLogger(Robot.class.getName()).log(Level.SEVERE, null, ex);}
    if(end==false){// not reach exit
       try{
            System.out.println(Thread.currentThread().getName());// displays current thread name
            System.out.println("pozisyon x="+x+"y="+y);
            decision(x,y); // thread makes decision to move
            visitedCell[x/40][y/40]+=1; // increade number of visitide cell count for that cell in the array
            visCell++; //increase visited cell count for the thread
            Thread.sleep(300);} // thread sleeps for a while to observe movement changing
       catch(Exception ex){
       }
    }
    else{// found exit
            Thread.currentThread().interrupt(); // Thread killed
            if(Main.butAct==2)// after a thread found exit, checks if result button can be active anymore
               Frame.button4.setEnabled(true); // activates result button
    }
  }//end while

}

multithreading

swing

java

回答 5

Stack Overflow用户

回答已采纳

发布于 2011-12-11 04:34:29

或者你可以这样做。manager类知道两个线程的位置，moveTo方法检查它们是否在同一位置重合。

class MazeManager {
  int x1, x2, y1, y2;

  public synchronized boolean moveTo(int threadId, int x, int y) {
    ..
  }
}

票数 1

Stack Overflow用户

发布于 2011-12-11 04:15:48

最简单的解决方案是以这样一种方式“划分”工作，即两个工作区域(要尝试的解决方案/路径)相互排斥。一种变通方法是拥有一个并发的位置集，每个线程在移动之前都会检查这些位置。

票数 1

Stack Overflow用户

发布于 2011-12-11 04:31:39

你将不得不使用所谓的互斥的概念。在Java语言中，您必须使用synchronized关键字来完成这项工作。可以在1中看到一个简单的示例

public class SynchronizedCounter {
  private int i = 0;

  public synchronized void increment() {
    i++;
  }

  public synchronized void decrement() {
    i--;
  }

  public synchronized int value() {
    return i;
  }
}

在这里您可以看到一段代码，它确保只有一个线程能够修改共享变量i的值。注意，代码使用this作为所谓的“锁对象”。您可以通过以下方式重写代码：

public class SynchronizedCounter {
  private int i = 0;

  public void increment() {
    synchronized(this) {
      i++;
    }
  }

  public void decrement() {
    synchronized(this) {
      i--;
    }
  }

  public int value() {
    synchronized(this) {
      return i;
    }
  }
}

或者，您可能希望创建自己的锁对象并使用它：

public class SynchronizedCounter {
  private int i = 0;
  private Object lock = new Object();

  public void increment() {
    synchronized(lock) {
      i++;
    }
  }

  public void decrement() {
    synchronized(lock) {
      i--;
    }
  }

  public int value() {
    synchronized(lock) {
      return i;
    }
  }
}

现在，任意数量的线程都可以随机调用这个对象的方法，但是一次只有一个线程能够传递锁并进行实际的修改。

票数 1

页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持

原文链接：

https://stackoverflow.com/questions/8459514

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