dwStruts 2011-05-20 09:45
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一个类似树要用多线程下执行怎么做,求解!

                                                                        A

                                          

 

                                                            B                C

                       

 

 

                                                                     D

 

 

 

 

                                                    E                 F             G

 

 

 

 

由于公司不能上传东西,也不能上其他网站;

现在想用多线程去执行,由于数据量很大,A下面有节点B 、C(如图),A执行完后才能执行B C,只有当B C两个节点都执行完了才能执行到D,如果B C有一个没执行完都不能执行D,只有当父节点执行完才能执行下面子节点,下面依次类推,必须用多线程,第一次发帖,大伙们不要笑话哦!

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  • fmjsjx 2011-05-20 11:12
    关注

    用文字讲太麻烦了,弄个简单的,楼主看看好了,个人感觉还可以改进。
    简单的代码示例:
    [code="java"]
    import java.util.Map;
    import java.util.concurrent.ConcurrentHashMap;

    public class MT {

    private final Map<String, Boolean> flags = new ConcurrentHashMap<String, Boolean>();
    
    private boolean hasDone(Operation... operations) {
        for (Operation operation : operations) {
            if (!flags.get(operation.getOperationName())) {
                return false;
            }
        }
        return true;
    }
    
    public static void main(String[] args) {
        MT mt = new MT();
        mt.flags.put("A", false);
        mt.flags.put("B", false);
        mt.flags.put("C", false);
        mt.flags.put("D", false);
        mt.A.run();
    }
    
    private Operation D = new Operation() {
    
        protected boolean isPrepareFinished() {
            return hasDone(D);
        }
    
        protected String getOperationName() {
            return "D";
        }
    
        protected void doing() {
            flags.put(getOperationName(), true);
        }
    };
    
    private Operation C = new Operation(D) {
    
        protected boolean isPrepareFinished() {
            return hasDone(B, C);
        }
    
        protected String getOperationName() {
            return "C";
        }
    
        protected void doing() {
            flags.put(getOperationName(), true);
        }
    };
    
    private Operation B = new Operation(D) {
    
        protected boolean isPrepareFinished() {
            return hasDone(B, C);
        }
    
        protected String getOperationName() {
            return "B";
        }
    
        protected void doing() {
            flags.put(getOperationName(), true);
        }
    };
    
    private Operation A = new Operation(B,C) {
    
        protected boolean isPrepareFinished() {
            return hasDone(A);
        }
    
        protected String getOperationName() {
            return "A";
        }
    
        protected void doing() {
            flags.put(getOperationName(), true);
        }
    };
    
    
    
    abstract class Operation implements Runnable {
    
        protected abstract String getOperationName();
    
        protected abstract void doing();
    
        protected abstract boolean isPrepareFinished();
    
        private Operation[] afterOperations;
    
        public Operation(Operation... afterOperations) {
            this.afterOperations = afterOperations;
        }
    
        public void run() {
            System.out.println("doing " + getOperationName() + " ...");
            doing();
            if (isPrepareFinished()) {
                Thread[] threads = new Thread[afterOperations.length];
                for (int i = 0; i < threads.length; i++) {
                    threads[i] = new Thread(afterOperations[i]);
                }
                for (Thread thread : threads) {
                    thread.run();
                }
            }
        }
    }
    

    }
    [/code]

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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