POJ 3624 Charm Bracelet （01背包）

CharmBracelet

TimeLimit:1000MS		MemoryLimit:65536K
TotalSubmissions:23950		Accepted:10802

Description

Bessiehasgonetothemall'sjewelrystoreandspiesacharmbracelet.Ofcourse,she'dliketofillitwiththebestcharmspossiblefromtheN(1≤N≤3,402)availablecharms.EachcharmiinthesuppliedlisthasaweightWi(1≤Wi≤400),a'desirability'factorDi(1≤Di≤100),andcanbeusedatmostonce.BessiecanonlysupportacharmbraceletwhoseweightisnomorethanM(1≤M≤12,880).

Giventhatweightlimitasaconstraintandalistofthecharmswiththeirweightsanddesirabilityrating,deducethemaximumpossiblesumofratings.

Input

*Line1:Twospace-separatedintegers:NandM
*Lines2..N+1:Linei+1describescharmiwithtwospace-separatedintegers:WiandDi

Output

*Line1:Asingleintegerthatisthegreatestsumofcharmdesirabilitiesthatcanbeachievedgiventheweightconstraints

SampleInput

46
14
26
312
27

SampleOutput

23

大致题意：贝西有一堆首饰，每个首饰有它的魅力值和重量。现在第一行输入n个首饰和最大重量为m。接下来n行每行两个数a和b，a为首饰的魅力值，b为该首饰的魅力值。求贝西能在不超过最大重量的情况下，最大魅力值是多少？

#include<iostream>
usingnamespacestd;
intf[12900];
intw[3410],v[3410];
intmain(void)
{
intn,m,i;
cin>>n>>m;
for(i=1;i<=n;i++)
cin>>w[i]>>v[i];
memset(f,0,sizeof(f));
for(i=1;i<=n;i++)
for(intj=m;j>=w[i];j--)
if(f[j-w[i]]+v[i]>f[j])
f[j]=f[j-w[i]]+v[i];
cout<<f[m]<<endl;
return0;
}