@zsh-o
2018-11-05T15:53:33.000000Z
字数 18417
阅读 1096
Java
首先给出一个同步问题的定义
这里就要提到python的动态类型的tuple的好处,可以实现统一形式的多输入和多输出
接下来从最简单的开始一步一步进行优化
首先从最基本的顺序结构执行,要保证结果的正确性,需要按照上述有向图的一种拓扑排序顺序来组织代码
顺序结构我们只能按照其一种拓扑排序的方法组织其运行,这里按照[A,B,C,D,E,F]顺序运行即可
package com.zsh_o.future;
/**
* 顺序结构执行
* */
public class SequentialBase {
public static void main(String[] args) {
SequentialBase app = new SequentialBase();
try {
app.run();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
void run() throws InterruptedException {
long startTime = System.currentTimeMillis();
int a = getA();
int b = getB(a);
int c = getC();
int d = getD();
int e = getE(c, d);
int f = getF(b, e);
long endTime = System.currentTimeMillis();
System.out.printf("Final Result: %d\n", f);
System.out.printf("Total Time: %d\n", endTime - startTime);
}
/**
* Define Functions A -> F
* */
int getA() throws InterruptedException {
System.out.println("A: Running");
Thread.sleep(1000);
System.out.println("A: Returned");
return 1;
}
int getB(int a) throws InterruptedException {
System.out.println("B: Running");
Thread.sleep(1000);
System.out.println("B: Returned");
return a + 1;
}
int getC() throws InterruptedException {
System.out.println("C: Running");
Thread.sleep(1000);
System.out.println("C: Returned");
return 10;
}
int getD() throws InterruptedException {
System.out.println("D: Running");
Thread.sleep(1000);
System.out.println("D: Returned");
return 20;
}
int getE(int c, int d) throws InterruptedException {
System.out.println("E: Running");
Thread.sleep(1000);
System.out.println("E: Returned");
return c + d;
}
int getF(int b, int e) throws InterruptedException {
System.out.println("F: Running");
Thread.sleep(1000);
System.out.println("F: Returned");
return b * e;
}
}
执行结果:
A: Running
A: Returned
B: Running
B: Returned
C: Running
C: Returned
D: Running
D: Returned
E: Running
E: Returned
F: Running
F: Returned
Final Result: 60
Total Time: 6003
可以看到执行效果还是非常不错的,加上输入输出也只多了3ms,再复杂的代码都是由最基础的代码一步一步按照场景优化来的,所以先来优化这个顺序结构
先从小功能开始,记录时间的代码太长了,如果要计时的太多会产生很多的计时变量,把其封装一下:
package com.zsh_o.util;
/**
* 用以记录程序运行时间
* */
public class CountTimer {
private long startTime;
private long endTime;
private long time;
public CountTimer() {
startTime = 0;
endTime = 0;
time = 0;
}
public void start() {
startTime = System.currentTimeMillis();
}
public void end() {
endTime = System.currentTimeMillis();
time = endTime - startTime;
}
public long getTime() {
return time;
}
}
第二个是,为了查看效果我们强行家了sleep一秒,但sleep是要catch异常的,所以每sleep一次都要处理下异常,在这个里面是完全没有必要的,我们单拿出来并处理下异常
package com.zsh_o.util;
public class Util {
public static void sleep(long ms) {
try {
Thread.sleep(ms);
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
接下来我们考虑如何对代码进行复用,首先发现可以抽象出函数的调用,这样用户只需要关心函数体的内容就可以了,所以抽象出函数的调用,增加Callable接口
package com.zsh_o.future;
public interface Callable<T> {
T call();
}
这个地方可以看到我们定义的接口没有传递参数,是由于我们不知道上层用户想要传递几个参数,也不知道每个参数是什么类型,所以最好的办法就是不写参数。。。
python里面传递参数相当于直接传递tuple和dict,然后在函数里面检验参数正确性,但对于Java这种静态编译语言这种方法太影响性能,直接用Object[]在函数里面类型转换,这样对函数参数的约束太弱,很容易崩溃。Java也用泛型实现Tuple,但无法保证长度,所以这个地方直接不加参数,用闭包实现功能。
package com.zsh_o.future;
import com.zsh_o.util.CountTimer;
import com.zsh_o.util.Util;
import java.util.ArrayList;
public class SequentialCase1 {
int a, b, c, d, e, f;
public static void main(String[] args) {
SequentialCase1 app = new SequentialCase1();
app.run();
}
void run() {
CountTimer timer = new CountTimer();
timer.start();
ArrayList<Callable> array = new ArrayList<>();
array.add(() -> {System.out.println("A: Running"); Util.sleep(1000); a = 1; System.out.println("A: Returned"); return a;});
array.add(() -> {System.out.println("B: Running"); Util.sleep(1000); b = a + 1; System.out.println("B: Returned"); return b;});
array.add(() -> {System.out.println("C: Running"); Util.sleep(1000); c = 10; System.out.println("C: Returned"); return c;});
array.add(() -> {System.out.println("D: Running"); Util.sleep(1000); d = 20; System.out.println("D: Returned"); return d;});
array.add(() -> {System.out.println("E: Running"); Util.sleep(1000); e = c + d; System.out.println("E: Returned"); return e;});
array.add(() -> {System.out.println("F: Running"); Util.sleep(1000); f = b * e; System.out.println("F: Returned"); return f;});
for(var e : array) {
e.call();
}
System.out.printf("Final Result: %d\n", f);
timer.end();
System.out.printf("Total Time: %d\n", timer.getTime());
}
}
结果如下
A: Running
A: Returned
B: Running
B: Returned
C: Running
C: Returned
D: Running
D: Returned
E: Running
E: Returned
F: Running
F: Returned
Final Result: 60
Total Time: 6046
可以看到ArrayList还是蛮耗时的
接下来我们发现,为了查看方法的执行效果,输出了每个方法的运行状态,但该运行状态在每个方法上都是相似的,包括sleep一秒也是相似的,所以,上面两个部分相当于对想实现功能的结果的附加和扩展,所以这里我们用装饰器模式对其扩展附加功能,首先定义该装饰器
package com.zsh_o.future;
/**
* 采用装饰器对Call进行装饰,为Call增加附加功能
* */
public class CallDecorator<T> implements Callable<T> {
Callable<T> callable;
String name;
public CallDecorator(String name, Callable<T> callable) {
this.callable = callable;
this.name = name;
}
@Override
public T call() {
System.out.printf("%s: Running\n", name);
try {
Thread.sleep(1000);
} catch (InterruptedException e) {
e.printStackTrace();
}
T value = callable.call();
System.out.printf("%s: Returned\n", name);
return value;
}
}
这里有一个问题,为什么是装饰器而不是代理?代理强调的是对代理对象的访问和执行控制,而装饰器强调对行为的扩展,这个地方只是加了输出和暂停一秒,所以叫装饰器更为合适
package com.zsh_o.future;
import com.zsh_o.util.CountTimer;
import java.util.ArrayList;
public class SequentialCase2{
int a, b, c, d, e, f;
public static void main(String[] args) {
SequentialCase2 app = new SequentialCase2();
try {
app.run();
} catch (InterruptedException e1) {
e1.printStackTrace();
}
}
void run() throws InterruptedException {
CountTimer timer = new CountTimer();
timer.start();
ArrayList<CallDecorator> array = new ArrayList<>();
array.add(new CallDecorator("A", () -> {a = 1; return a;}));
array.add(new CallDecorator("B", () -> {b = a + 1; return b;}));
array.add(new CallDecorator("C", () -> {c = 10; return c;}));
array.add(new CallDecorator("D", () -> {d = 20; return d;}));
array.add(new CallDecorator("E", () -> {e = c + d; return e;}));
array.add(new CallDecorator("F", () -> {f = b * e; return f;}));
for(var e : array) {
e.call();
}
timer.end();
System.out.printf("Final Result: %d\n", f);
System.out.printf("Total Time: %d\n", timer.getTime());
}
}
结果:
A: Running
A: Returned
B: Running
B: Returned
C: Running
C: Returned
D: Running
D: Returned
E: Running
E: Returned
F: Running
F: Returned
Final Result: 60
Total Time: 6055
我们用并发的方式优化该代码,该程序理论上最少的运行时间是3s,每一列的代码可以一起执行。在最一开始我们采用了6把锁来完成这个同步问题
package com.zsh_o.future;
import com.zsh_o.util.CountTimer;
import com.zsh_o.util.Util;
import java.util.ArrayList;
import java.util.concurrent.CountDownLatch;
import java.util.concurrent.locks.ReentrantLock;
/**
* 多线程解决,可能会出现的问题是还没来得及加锁,其他的线程已经开始运行
* */
public class ThreadBase1 {
private int a, b, c, d, e, f;
private ReentrantLock lockA, lockB, lockC, lockD, lockE, lockF;
public ThreadBase1() {
this.lockA = new ReentrantLock();
this.lockB = new ReentrantLock();
this.lockC = new ReentrantLock();
this.lockD = new ReentrantLock();
this.lockE = new ReentrantLock();
this.lockF = new ReentrantLock();
}
void run() throws InterruptedException {
CountTimer timer = new CountTimer();
timer.start();
// 用CountDownLatch确保主线程在所有子线程执行完再执行
CountDownLatch latch = new CountDownLatch(6);
ArrayList<Thread> array = new ArrayList<>();
array.add(new Thread(() -> {
lockA.lock();
System.out.println("A: Running");
Util.sleep(1000);a = 1;
System.out.println("A: Returned");
lockA.unlock();
latch.countDown();
}));
array.add(new Thread(() -> {
lockB.lock();lockA.lock();
System.out.println("B: Running");
Util.sleep(1000);b = a + 1;
System.out.println("B: Returned");
lockA.unlock();lockB.unlock();
latch.countDown();
}));
array.add(new Thread(() -> {
lockC.lock();
System.out.println("C: Running");
Util.sleep(1000);c = 10;
System.out.println("C: Returned");
lockC.unlock();
latch.countDown();
}));
array.add(new Thread(() -> {
lockD.lock();
System.out.println("D: Running");
Util.sleep(1000);d = 20;
System.out.println("D: Returned");
lockD.unlock();
latch.countDown();
}));
array.add(new Thread(() -> {
lockE.lock();lockC.lock();lockD.lock();
System.out.println("E: Running");
Util.sleep(1000);e = c + d;
System.out.println("E: Returned");
lockD.unlock();lockC.unlock();lockE.unlock();
latch.countDown();
}));
array.add(new Thread(() -> {
lockF.lock();lockB.lock();lockE.lock();
System.out.println("F: Running");
Util.sleep(1000);f = b * e;
System.out.println("F: Returned");
lockE.unlock();lockB.unlock();lockF.unlock();
latch.countDown();
}));
for(var e: array) {
e.start();
}
latch.await();
timer.end();
System.out.printf("Final Result: %d\n", f);
System.out.printf("Total Time: %d\n", timer.getTime());
}
public static void main(String[] args) {
ThreadBase1 app = new ThreadBase1();
try {
app.run();
} catch (InterruptedException e1) {
e1.printStackTrace();
}
}
}
结果:
D: Running
A: Running
C: Running
D: Returned
C: Returned
A: Returned
E: Running
B: Running
E: Returned
B: Returned
F: Running
F: Returned
Final Result: 60
Total Time: 3030
这个可能会出现的问题是,可能会出现还没来得及加锁后面的就运行了,则会出现错误结果,所以我们用CountDownLatch来修改该代码
package com.zsh_o.future;
import com.zsh_o.util.CountTimer;
import com.zsh_o.util.Util;
import java.util.ArrayList;
/**
* 多线程解决
* 用CountDownLatch完善
* */
public class ThreadCase1 {
private int a, b, c, d, e, f;
private CountDownLatch latchB, latchE, latchF;
public ThreadCase1() {
latchB = new CountDownLatch(1);
latchE = new CountDownLatch(2);
latchF = new CountDownLatch(2);
}
void run() throws InterruptedException {
CountTimer timer = new CountTimer();
timer.start();
CountDownLatch latchMain = new CountDownLatch(1);
ArrayList<Thread> array = new ArrayList<>();
array.add(new Thread(() -> {
System.out.println("A: Running");
Util.sleep(1000);a = 1;
System.out.println("A: Returned");
latchB.countDown();
}));
array.add(new Thread(() -> {
try {
latchB.await();
System.out.println("B: Running");
Util.sleep(1000);b = a + 1;
System.out.println("B: Returned");
latchF.countDown();
} catch (InterruptedException e1) {
e1.printStackTrace();
}
}));
array.add(new Thread(() -> {
System.out.println("C: Running");
Util.sleep(1000);c = 10;
System.out.println("C: Returned");
latchE.countDown();
}));
array.add(new Thread(() -> {
System.out.println("D: Running");
Util.sleep(1000);d = 20;
System.out.println("D: Returned");
latchE.countDown();
}));
array.add(new Thread(() -> {
try {
latchE.await();
System.out.println("E: Running");
Util.sleep(1000);e = c + d;
System.out.println("E: Returned");
latchF.countDown();
} catch (InterruptedException e1) {
e1.printStackTrace();
}
}));
array.add(new Thread(() -> {
try {
latchF.await();
System.out.println("F: Running");
Util.sleep(1000);f = b * e;
System.out.println("F: Returned");
latchMain.countDown();
} catch (InterruptedException e1) {
e1.printStackTrace();
}
}));
for(var e: array) {
e.start();
}
latchMain.await();
System.out.printf("Final Result: %d\n", f);
timer.end();
System.out.printf("Total Time: %d\n", timer.getTime());
}
public static void main(String[] args) {
ThreadCase1 app = new ThreadCase1();
try {
app.run();
} catch (InterruptedException e1) {
e1.printStackTrace();
}
}
}
执行结果与上面相同,可以看到CountownLatch非常好用,而且原理非常简单,那么我们来实现个吧
原理:await()阻断线程,并且MaxCount个线程执行countDown()之后才打开阻断,需要是不同的线程,需要注意的是countDown()函数不截断执行的线程,只截断执行await()的线程,如果想同时截断需要配合使用
package com.zsh_o.future;
import com.zsh_o.util.CountTimer;
import com.zsh_o.util.Util;
import javax.swing.event.MouseInputListener;
import java.util.concurrent.locks.ReentrantLock;
public class CountDownLatch {
private int count;
final private Object gobal;
private ReentrantLock lockCount;
public CountDownLatch(int count) {
this.count = count;
lockCount = new ReentrantLock();
gobal = new Object();
}
public void countDown() {
lockCount.lock();
if(count > 0)
count--;
else return;
if(count == 0)
synchronized (gobal) {
gobal.notifyAll();
}
lockCount.unlock();
}
public void await() throws InterruptedException {
synchronized (gobal) {
if (count > 0)
gobal.wait();
}
}
}
这里用了java的Object的wait和noifyAll()函数,用以任意数量的线程等待和恢复
上面的问题是需要用户自己来考虑同步机制,如何把同步方法也封装到代码里面,让用户只关心代码逻辑即可,这个地方用的是Future模式,顾名思义,Future表示的是未来的一种意愿,表示未来会发生的事情,这个地方表示该实体在未来会得到返回值,调用future.get的表示期望得到给结果,如果结果还没有准备好则等待结果,例如上面的E会执行int e = futureC.get() + futureD.get()则E会等待C和D执行完之后继续执行。这个地方涉及到两个点:一,如何进入等待,二,如何从等待中恢复;这里的原则是等待过程不能占用计算资源,也就是不能用while一直判断的方法实现,那么实现方法就应该是事件驱动,进入等待释放计算资源,从等待恢复再重新可计算。
这个地方需要用代理模式,用代理形成逻辑,最后具体的执行由代理控制
首先定义futurable接口
package com.zsh_o.future;
public interface Futurable<T> {
T get();
}
下面是Future抽象类
package com.zsh_o.future;
public abstract class Future<T> implements Futurable<T>, Runnable {
protected Callable<T> callable;
protected boolean finished;
public Future() {
this.callable = null;
this.finished = false;
}
public void register(Callable<T> callable) {
this.callable = callable;
}
}
接下来是以线程的方式实现Future,当然也可以实现分布式的方式,核心思想相似
package com.zsh_o.future;
import com.zsh_o.util.CountTimer;
import com.zsh_o.util.Util;
public class ThreadFuture<T> extends Future<T> {
private CountDownLatch latch;
private T value;
private boolean finished;
private final Object lock;
public ThreadFuture() {
latch = new CountDownLatch(1);
value = null;
finished = false;
lock = new Object();
}
@Override
public T get() {
try {
if (callable == null)
throw new Exception("Unregistered Callable");
if (!finished){
run();
latch.await();
}
} catch (Exception e) {
e.printStackTrace();
}
return value;
}
@Override
public void run() {
new Thread(()->{
synchronized (lock) {
if (!finished) {
value = callable.call();
finished = true;
latch.countDown();
}
}
}).start();
}
public void reset() {
synchronized (lock) {
finished = false;
}
}
public boolean getState() {
return finished;
}
}
实现的功能是,用register注册函数体逻辑Callable,需要注意的是Callable中需要调用代理的future.get()方法,以形成执行逻辑,get()执行await进入等待,等待其他线程执行完run(),并且保证只执行一次
package com.zsh_o.future;
import com.zsh_o.util.CountTimer;
import com.zsh_o.util.Util;
public class ThreadCase2 {
private ThreadFuture<Integer> futureA, futureB, futureC, futureD, futureE, futureF;
public ThreadCase2() {
this.futureA = new ThreadFuture<>();
this.futureB = new ThreadFuture<>();
this.futureC = new ThreadFuture<>();
this.futureD = new ThreadFuture<>();
this.futureE = new ThreadFuture<>();
this.futureF = new ThreadFuture<>();
}
void run() {
CountTimer timer = new CountTimer();
timer.start();
futureA.register(()->{System.out.println("A: Runing");int a = 1;Util.sleep(1000);System.out.println("A: Over");return a;});
futureB.register(()->{System.out.println("B: Runing");int b = futureA.get()+1;Util.sleep(1000);System.out.println("B: Over");return b;});
futureC.register(()->{System.out.println("C: Runing");int c = 10;Util.sleep(1000);System.out.println("C: Over");return c;});
futureD.register(()->{System.out.println("D: Runing");int d = 20;Util.sleep(1000);System.out.println("D: Over");return d;});
futureE.register(()->{System.out.println("E: Runing");int e = futureC.get()+futureD.get();Util.sleep(1000);System.out.println("E: Over");return e;});
futureF.register(()->{System.out.println("F: Runing");int f = futureB.get()*futureE.get();Util.sleep(1000);System.out.println("F: Over");return f;});
futureA.run();futureB.run();futureC.run();futureD.run();futureE.run();futureF.run();
System.out.printf("Final Result: %d\n", futureF.get());
timer.end();
System.out.printf("Total Time: %d\n", timer.getTime());
}
public static void main(String[] args) {
ThreadCase2 app = new ThreadCase2();
app.run();
}
}
结果:
A: Runing
B: Runing
C: Runing
D: Runing
E: Runing
F: Runing
C: Over
D: Over
A: Over
E: Over
B: Over
F: Over
Final Result: 60
Total Time: 3051
最后,如果并发的非常多可能会崩掉,所以我们加了一个限制线程执行状态最大值的方法
package com.zsh_o.future;
import java.util.concurrent.ConcurrentHashMap;
import java.util.concurrent.ConcurrentLinkedDeque;
public class ThreadPoolFuture<T> {
private class Item<D> extends Future<D> {
Thread thread;
D value;
CountDownLatch latch;
final Object lock;
boolean finished;
public Item() {
this.value = null;
this.latch = new CountDownLatch(1);
this.lock = new Object();
this.finished = false;
}
@Override
public D get() {
return value;
}
@Override
public void run() {
if (!thread.isAlive()) {
thread.start();
}
}
}
private int maxPool;
private final Object lockObject;
private ConcurrentHashMap<String, Item<T>> pool;
private ConcurrentLinkedDeque<Item<T>> poolList;
private int currentRuning;
private CountDownLatch lathGobal;
public ThreadPoolFuture(int maxPool) {
this.maxPool = maxPool;
this.pool = new ConcurrentHashMap<>();
poolList = new ConcurrentLinkedDeque<>();
this.lockObject = new Object();
this.currentRuning = 0;
}
public void register(String name, Callable<T> callable) {
Item<T> item = new Item<>();
item.thread = new Thread(()->{
synchronized (item.lock) {
if (!item.finished) {
synchronized (lockObject) {
while (currentRuning >= maxPool) {
try {
lockObject.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
currentRuning++;
System.out.println(name + " Run: " + currentRuning);
}
item.value = callable.call();
synchronized (lockObject) {
currentRuning--;
lockObject.notifyAll();
}
item.latch.countDown();
lathGobal.countDown();
item.finished = true;
}
}
});
pool.put(name, item);
poolList.push(item);
}
public T get(String name) {
if (!pool.containsKey(name)) {
try {
throw new Exception("Unregister Callable");
} catch (Exception e) {
e.printStackTrace();
}
}
Item<T> item = pool.get(name);
if (!item.finished) {
try {
synchronized (lockObject) {
lockObject.notifyAll();
currentRuning--;
System.out.println(name + " Get: " + currentRuning);
}
item.run();
item.latch.await();
synchronized (lockObject) {
while (currentRuning >= maxPool) {
try {
lockObject.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
currentRuning++;
System.out.println(name + " Get: " + currentRuning);
}
} catch (InterruptedException e) {
e.printStackTrace();
return null;
}
}
return item.value;
}
public void start() {
lathGobal = new CountDownLatch(poolList.size());
for(var e: poolList) {
e.run();
}
}
public void await() throws InterruptedException {
if (lathGobal == null)
lathGobal = new CountDownLatch(poolList.size());
lathGobal.await();
}
}
基本思路是,由于上面我们用消息的方法执行等待和恢复执行,所以进入等待和恢复执行对于我们来说是可见的,所以只需要监控当前运行的最大线程即可,只是执行状态的数量不包括阻塞和等待
package com.zsh_o.future;
import com.zsh_o.util.CountTimer;
import com.zsh_o.util.Util;
/**
* Future模式解决问题,限制最大运行线程数
* */
public class ThreadCase3 {
private ThreadPoolFuture<Integer> pool;
public ThreadCase3() {
this.pool = new ThreadPoolFuture<>(2);
}
void run() throws Exception {
CountTimer timer = new CountTimer();
timer.start();
pool.register("A", ()->{
System.out.println("A: Running");
int a = 1;
Util.sleep(1000);
System.out.println("A: Over");
return a;
});
pool.register("B", ()->{
System.out.println("B: Running");
int b = pool.get("A") + 1;
Util.sleep(1000);
System.out.println("B: Over");
return b;
});
pool.register("C", ()->{
System.out.println("C: Running");
int c = 10;
Util.sleep(1000);
System.out.println("C: Over");
return c;
});
pool.register("D", ()->{
System.out.println("D: Running");
Util.sleep(1000);
System.out.println("D: Over");
return 20;
});
pool.register("E",()->{
System.out.println("E: Running");
Util.sleep(1000);
int e = pool.get("C") + pool.get("D");
System.out.println("E: Over");
return e;
});
pool.register("F", ()->{
System.out.println("F: Running");
int f = pool.get("B") * pool.get("E");
Util.sleep(1000);
System.out.println("F: Over");
return f;
});
pool.start();
pool.await();
System.out.printf("Final Result: %d\n", pool.get("F"));
timer.end();
System.out.printf("Total Time: %d\n", timer.getTime());
}
public static void main(String[] args) {
ThreadCase3 app = new ThreadCase3();
try {
app.run();
} catch (Exception e) {
e.printStackTrace();
}
}
}
结果:
maxCount = 2
F Run: 1
F: Running
B Get: 0
A Run: 1
A: Running
B Run: 2
B: Running
A Get: 1
C Run: 2
C: Running
A: Over
C: Over
E Run: 2
E: Running
D Run: 2
D: Running
D Get: 1
D: Over
A Get: 2
D Get: 2
E: Over
B: Over
B Get: 1
F: Over
Final Result: 60
Total Time: 4072
------------------------------
maxCount = 1
F Run: 1
F: Running
B Get: 0
A Run: 1
A: Running
A: Over
E Run: 1
E: Running
C Get: 0
B Run: 1
B: Running
B: Over
D Run: 1
D: Running
D: Over
B Get: 1
E Get: 0
C Run: 1
C: Running
C: Over
C Get: 1
E: Over
E Get: 1
F: Over
Final Result: 60
Total Time: 6098
---------------------
maxCount = 3
F Run: 1
F: Running
A Run: 2
A: Running
B Run: 3
B: Running
A Get: 2
E Run: 3
E: Running
B Get: 2
D Run: 3
D: Running
A: Over
C Run: 3
D: Over
C Get: 2
C: Running
A Get: 3
C: Over
C Get: 2
E: Over
B: Over
B Get: 1
F: Over
Final Result: 60
Total Time: 3088
-------------------------
maxCount = 4
F Run: 1
F: Running
A Run: 2
A: Running
B Run: 3
B: Running
C Run: 4
C: Running
A Get: 3
B Get: 2
E Run: 3
E: Running
D Run: 4
D: Running
A: Over
C: Over
D: Over
C Get: 2
A Get: 2
C Get: 2
E: Over
B: Over
B Get: 1
F: Over
Final Result: 60
Total Time: 3078