这期内容当中小编将会给大家带来有关用coffee和socket.io实现的01背包算法是怎样的,文章内容丰富且以专业的角度为大家分析和叙述,阅读完这篇文章希望大家可以有所收获。
先说说我为什么写这些吧
当程序猿太苦逼了,真的,时间久了,真没有搬砖的成就感高,好歹人家能盖栋楼(身材也能练得不错),咱们指不定哪天来个熊孩子把硬盘格了就啥也没了。
这学期明显没把心放在前端上……汗啊,将来还想吃着口饭呢,但是这学期绝对没休息,只是忙了很多可能很多人认为无聊的事。
因为这学期无聊事太多了,耽误了很多,也让导师很失望,自己也很自卑,整理一下调调心态。
因为很多是针对作业的奇葩想法,所以,作业嘛,不糊弄就不是作业了,还希望大家多多批评。
兴许因为哪篇文章能解决工作呢。
我想试试Markdown。
靓照一张
进入正题
后台实现部分:
io = require “socket.io” http = require “http” fs = require “fs” express = require “express” mime = require “mime” app = express() server = http.createServer app server.listen 8080 console.log “Listening 8080”
app.get “/“,(req,res)->
path = "#{__dirname}/console.html" res.writeHead 200,"Content-Type":mime.lookup(path) res.end fs.readFileSync path
app.get “/jquery.min.js”,(req,res)->
path = "#{__dirname}/jquery.min.js" res.writeHead 200,"Content-Type":mime.lookup(path) res.end fs.readFileSync path
app.get “/bootstrap.min.js”,(req,res)->
path = "#{__dirname}/bootstrap.min.js" res.writeHead 200,"Content-Type":mime.lookup(path) res.end fs.readFileSync path
app.get “/bootstrap.min.css”,(req,res)->
path = "#{__dirname}/bootstrap.min.css" res.writeHead 200,"Content-Type":mime.lookup(path) res.end fs.readFileSync path
getCurrentTime = ->
d = new Date()
return “#{d.getFullYear()}-#{d.getMonth()+1}-#{d.getDate()} #{d.getHours()}:#{d.getMinutes()}:#{d.getSeconds()}”
class dynamicPack
pack:(data)-> c=[] i=0 j=0 while i<data.m+1 c[i]=[] c[i][0]=0 i++ while j<data.n+1 c[0][j]=0 j++ i=1 while i<data.m+1 j=1 while j<data.n+1 if data.w[i-1]<=j if c[i-1][j]<c[i-1][j-data.w[i-1]]+data.v[i-1] c[i][j]=c[i-1][j-data.w[i-1]]+data.v[i-1] else c[i][j]=c[i-1][j] else c[i][j] = c[i-1][j] j++ i++ return c; print:(c,data)-> x = [] i = data.m n = data.n str = "" #console.log c[i][m] while i>0 if c[i][n] > c[i-1][n] x[i-1] = 1 n -= data.w[i-1] else x[i-1] = 0 i-- i= 0 count = 0 while i<data.m count += x[i]*data.v[i] str += (i+1)+"," if x[i]!=0 i++ return str+"共计价值#{count}"
class knapPack
pack : (data)-> @v = data.v @w = data.w @m = data.m @n = data.n @cw = 0 @cv = 0 @put = [] @bestp = 0 temp_order = 0; temp = 0 perp = [] i=0 while i<@m perp[i] = @v[i]/@w[i] @put[i] = 0; i++ console.log perp i=0 while i<@m j=i+1 while j<@m if perp[i]<perp[j] temp = @v[i] @v[i] = @v[j] @v[j] = temp temp = @w[i] @w[i] = @w[j] @w[j] = temp j++ i++ backtrack : (i)-> console.log i @bound i if i>@m @bestp = @cv return if @cw+@w[i]<=@n @cw+=@w[i] @cv+=@v[i] @put[i]=1 @backtrack(i+1) @cw-=@w[i] @cv-=@v[i] if @bound(i+1)>@bestp @backtrack(i+1) bound :(i)-> leftw = @n - @cw b = @cv while i<=@m and @w[i]<=leftw leftw -= @w[i] b += @v[i] i++ b+=@v[i]/@w[i]*leftw if i<@m return b print :(data)-> @pack(data) console.log @w console.log @v @backtrack(0) console.log @put return @bestp
dask = (msg)->
answer = "" data = JSON.parse msg console.log data d = new dynamicPack() console.log d.pack(data) answer += "动态规划,选择物品"+d.print d.pack(data),data return answer
kask = (msg)->
answer = "" data = JSON.parse msg console.log data k = new knapPack() answer += "分支限界,***解"+k.print data return answer
io.listen(server).on “connection”,(socket)->
socket.on "msg",(msg)-> ##console.log msg socket.emit "msg",{time:getCurrentTime(),text:"calculating..."} socket.emit "msg",{time:getCurrentTime(),text:dask(msg)} socket.emit "msg",{time:getCurrentTime(),text:kask(msg)} ##socket.broadcast.emit "msg",data console.log "#{getCurrentTime()}:Connected"
前端实现部分:
输入示例:{"n":10,"m":3,"w":[3,4,5],"v":[4,5,6]}其中n为背包容量,m为物品数量
上述就是小编为大家分享的用coffee和socket.io实现的01背包算法是怎样的了,如果刚好有类似的疑惑,不妨参照上述分析进行理解。如果想知道更多相关知识,欢迎关注编程网行业资讯频道。