Charm Bracelet poj 3624 （01） 背包问题 c++

Charm Bracelet

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 23076 Accepted: 10383

Description

Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from theN (1 ≤N ≤ 3,402) available charms. Each charm i in the supplied list has a weightW_i (1 ≤W_i ≤ 400), a 'desirability' factorD_i (1 ≤D_i ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more thanM (1 ≤M ≤ 12,880).

Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: Line i+1 describes charm i with two space-separated integers:W_i andD_i

Output

* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints

Sample Input

4 61 42 63 122 7

Sample Output

23

#include<cstdio>#include<cstring>#include<cmath>using  namespace  std;int w[5000],v[5000];int f[14000];int max(int a,int b){    if(a<b)       return b;    return a;}int main(){    int i,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]);             }    return 0;}

 

 

//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>int main(){    int a,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]);    }    return 0;}