一个小学题目的解: 规则引擎Drools实现

最新推荐文章于 2022-08-26 14:28:07 发布

iteye_19871

最新推荐文章于 2022-08-26 14:28:07 发布

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本文通过使用规则引擎Drools实现了一个经典的数学问题——汽水瓶问题的解决方案。小明利用初始资金购买汽水并利用空瓶换汽水，最终最大化汽水饮用数量。文章详细介绍了Drools规则的定义及程序实现过程。

同学帮他侄儿问我一个问题：
资源：
1、小明的钱可以购买50瓶汽水。
2、老板搞促销，两个空汽水瓶子可以换一瓶汽水。

问：小明最多能喝多少瓶汽水？

开始还楞了一下，现在的小学题目还真不简单。不过马上给出答案：
先是50瓶，然后用50个空瓶换来25瓶，喝完后用25个空瓶换12瓶再多个空瓶，
喝完后用13个空瓶换6瓶，然后是7个空瓶换3瓶，然后是4个空瓶换2瓶，
然后是2个空瓶换1瓶，最后问老板借1个空瓶，再用2个空瓶换1瓶，
剩下来那个空瓶还给老板。所以喝到50+25+12+6+3+2+1+1=100

不过最近在学习规则引擎，自然就想到了怎么用规则引擎来实现。
我的想法是：
1: 假设汽水一元一瓶，则小明最初由50元。
2: 如果兜里有超过一元钱，则买一瓶汽水喝，此时：
钱减少一元，同时拥有的空瓶增加一个。
3: 如果拥有至少两个空瓶，则两个空瓶卖给老板，兜里的钱加一。
根据如上想法，有了如下Drool的实现。

java文件SodaWater.java：

packageorg.drools.examples;

importjava.io.InputStreamReader;

importjava.io.Reader;

importorg.drools.FactHandle;

importorg.drools.RuleBase;

importorg.drools.RuleBaseFactory;

importorg.drools.StatefulSession;

importorg.drools.WorkingMemory;

importorg.drools.compiler.PackageBuilder;

importorg.drools.rule.Package;

publicclassSodaWater
{

publicstaticfinalvoidmain(String[]args)throwsException{

finalPackageBuilderbuilder=newPackageBuilder();

builder.addPackageFromDrl(newInputStreamReader(SodaWater.class.getResourceAsStream("SodaWater.drl")));

finalRuleBaseruleBase=RuleBaseFactory.newRuleBase();
ruleBase.addPackage(builder.getPackage());

finalStatefulSessionsession=ruleBase.newStatefulSession();

Customercustomer=newCustomer("XiaoMing",50);
session.insert(customer);
session.fireAllRules();
}

publicstaticclassCustomer
{

privateStringname;

privateintmoney;

privateintdrinkSum;

privateintblankCup;

publicCustomer(Stringname,intmoney)
{

this.name=name;

this.money=money;

this.drinkSum=0;

this.blankCup=0;
}

publicintgetMoney()
{

returnmoney;
}

publicvoidsetMoney(intmoney)
{

this.money=money;
}

publicintgetBlankCup()
{

returnblankCup;
}

publicvoidsetBlankCup(intblankCup)
{

this.blankCup=blankCup;
}

publicintgetDrinkSum()
{

returndrinkSum;
}

publicvoidsetDrinkSum(intdrinkSum)
{

this.drinkSum=drinkSum;
}

publicStringgetName()
{

returnname;
}

publicvoidsetName(Stringname)
{

this.name=name;
}
}
}

rule文件：

packageorg.drools.examples

dialect"mvel"

importorg.drools.examples.SodaWater.Customer

rule"buyasodawateranddrink"
when

$c:Customer(money>0,$m:money,$b:blankCup,$d:drinkSum)
then

$c.money=$m-1;

$c.blankCup=$b+1;

$c.drinkSum=$d+1;

System.out.println("Customer"+$c.name+"nowbuyasodawateranddrink:money="+$c.money+"andblankCup="+$c.blankCup);
update($c);
end

rule"saleblankcupandgetmoney"
when

$c:Customer(blankCup>1,$b:blankCup,$m:money)
then

$c.blankCup=$b-2;

$c.money=$m+1;

System.out.println("Customer"+$c.name+"nowsale2cupsandgetmoney:money="+$c.money+",blankCup="+$c.blankCup);
update($c);
end

rule"finishdrink"

no-looptrue

dialect"java"
when

$c:Customer(blankCup<2,money==0)
then

System.out.println("Customer"+$c.getName()+"finisheddrink,anddrinknumberis"+$c.getDrinkSum()+"blankCup="+$c.getBlankCup());
end

运行结果如下：
Customer XiaoMing now buy a soda water and drink: money=49 and blankCup=1
Customer XiaoMing now buy a soda water and drink: money=48 and blankCup=2
Customer XiaoMing now sale 2 cups and get money: money=49, blankCup=0
Customer XiaoMing now buy a soda water and drink: money=48 and blankCup=1
Customer XiaoMing now buy a soda water and drink: money=47 and blankCup=2

...

Customer XiaoMing now buy a soda water and drink: money=1 and blankCup=1
Customer XiaoMing now buy a soda water and drink: money=0 and blankCup=2
Customer XiaoMing now sale 2 cups and get money: money=1, blankCup=0
Customer XiaoMing now buy a soda water and drink: money=0 and blankCup=1
Customer XiaoMing finished drink, and drink number is 99 blankCup=1

不过小明最后手上还有最后一个瓶子，计算机只能到此了。
向老板借一个瓶子换一瓶汽水，喝完在还一个空瓶子，估计就是人类智慧来。

此程序需要drools的jar包，具体可见：

http://lcllcl987.iteye.com/blog/254381