POJ 3624 Charm Bracelet

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 the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight Wi (1 ≤ Wi ≤ 400), a 'desirability' factor Di (1 ≤ Di ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (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: Wi and Di

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 6
1 4
2 6
3 12
2 7

Sample Output

23
01背包详见背包九讲:http://blog.csdn.net/ultimater/article/details/7769429
LANGUAGE:C
CODE:
#include<stdio.h>
#define max(a,b) a>b?a:b
int main()
{
    //freopen("in.txt","r",stdin);
    int n,v,i,j,f[12900]={0},get[3405],cost[3405];
    scanf("%d%d",&n,&v);
    for(i=1;i<=n;i++)
        scanf("%d%d",&cost[i],&get[i]);
    for(i=1;i<=n;i++)
        for(j=v;j>=cost[i];j--)
            f[j]=max(f[j],f[j-cost[i]]+get[i]);
    printf("%d\n",f[v]);
    return 0;
}


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