Charm Bracelet poj 3624 （01） 背包问题 c++

CharmBracelet

TimeLimit:1000MS		MemoryLimit:65536K
TotalSubmissions:23076		Accepted:10383

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:Wiand
Di

Output

*Line1:Asingleintegerthatisthegreatestsumofcharmdesirabilitiesthatcanbeachievedgiventheweightconstraints

SampleInput

46
14
26
312
27

SampleOutput

23

#include<cstdio>
#include<cstring>
#include<cmath>
usingnamespacestd;

intw[5000],v[5000];
intf[14000];

intmax(inta,intb)
{
if(a<b)
returnb;
returna;
}

intmain()
{
inti,j,n,m;

while(scanf("%d%d",&n,&m)!=EOF)
{
memset(w,0,sizeof(w));
memset(v,0,sizeof(v));
memset(f,0,sizeof(f));
for(i=1;i<=n;i++)
scanf("%d%d",&w[i],&v[i]);
for(i=1;i<=n;i++)
for(j=m;j>=w[i]&&j>0;j--)
f[j]=max(f[j],f[j-w[i]]+v[i]);
printf("%d\n",f[m]);
}
return0;
}

//第一次做01背包问题，是个入门题，注意数组开的要大一点，还有就是，这个题用二维数组会超内存；

后来和同学讨论的只定义一个数组就可以了，没必要定义三个，代码如下：

#include<cstdio>
#include<cstring>
intmain()
{
inta,b,dp[1100],i,j,v,n;
while(scanf("%d%d",&n,&v),!(n==0&&v==0))
{
memset(dp,0,4400);
for(i=1;i<=n;i++)
{
scanf("%d%d",&a,&b);
for(j=v;j>=a;j--)
if(dp[j]<dp[j-a]+b)
dp[j]=dp[j-a]+b;
}
printf("%d\n",dp[v]);
}
return0;
}