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文件名称:diaodu

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  • 上传时间:
    2012-11-16
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    13.84kb
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两台处理机A 和B处理n个作业。设第i个作业交给机器

A 处理时需要时间ai,若由机器B 来处理,则需要时间bi。由于各作

业的特点和机器的性能关系,很可能对于某些i,有ai >=bi,而对于

某些j,j!=i,有aj<bj。既不能将一个作业分开由两台机器处理,也没

有一台机器能同时处理2 个作业。设计一个动态规划算法,使得这两

台机器处理完成这n 个作业的时间最短(从任何一台机器开工到最后

一台机器停工的总时间)。研究一个实例:(a1,a2,a3,a4,a5,a6)=

(2,5,7,10,5,2);(b1,b2,b3,b4,b5,b6)=(3,8,4,11,3,4)-Two processors A and B deal with n-operation. I set up the first operating the machine A treatment takes time ai, if by machine B to deal with, it takes time bi. Since the operating characteristics and performance of the machine relationship, is likely to for some i, there is ai> = bi, and for some j, j! = I, there is aj <bj. Neither can an operation separate from the two machines to deal with, nor a machine capable of simultaneously handling two operations. Design a dynamic programming algorithm, making the two machines to deal with the completion of this n the shortest operating time (from any one machine starts to the last machine downtime total time). Study an example of: (a1, a2, a3, a4, a5, a6) = (2,5,7,10,5,2) (b1, b2, b3, b4, b5, b6) = (3,8 , 4,11,3,4)
(系统自动生成,下载前可以参看下载内容)

下载文件列表

diaodu/diaodu.jpx.local
diaodu/diaodu.jpx
diaodu/src/package.html
diaodu/src/diaodu/dd.java
diaodu/src/diaodu/dd.jbx
diaodu/src/diaodu
diaodu/src
diaodu/bak/package.html~1~
diaodu/bak/diaodu/dd.java~1~
diaodu/bak/diaodu/dd.java~2~
diaodu/bak/diaodu/dd.java~3~
diaodu/bak/diaodu/dd.java~4~
diaodu/bak/diaodu/Element.java~1~
diaodu/bak/diaodu/Element.java~2~
diaodu/bak/diaodu/dd.java~5~
diaodu/bak/diaodu/dd.java~6~
diaodu/bak/diaodu/dd.java~7~
diaodu/bak/diaodu/dd.java~8~
diaodu/bak/diaodu/dd.java~9~
diaodu/bak/diaodu/dd.java~10~
diaodu/bak/diaodu
diaodu/bak
diaodu/classes/diaodu/dd.class
diaodu/classes/diaodu/dd$Element.class
diaodu/classes/diaodu
diaodu/classes/package cache/diaodu.dep2
diaodu/classes/package cache
diaodu/classes
diaodu/diaodu.oif
diaodu/doc/diaodu/dd.html
diaodu/doc/diaodu
diaodu/doc
diaodu

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