poj3624

Charm Bracelet

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 40383 Accepted: 17532

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 weightW_i (1 ≤ W_i ≤ 400), a 'desirability' factor D_i (1 ≤ D_i ≤ 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: W_i and D_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

AC代码：

#include<iostream>
#include<algorithm>
#include<string.h>
#include<stdio.h>
using namespace std;
int main()
{
        int n,m;
        while(scanf("%d%d",&n,&m)!=EOF)
        {
                int w[3500],d[13000];
                int f[13000];
                for(int i=1;i<=n;i++)
                {
                        scanf("%d%d",&w[i],&d[i]);
                }
                for(int i=1;i<=n;i++)
                {
                        for(int j=m;j>=w[i];j--)
                                f[j]=max(f[j],f[j-w[i]]+d[i]);
                }
                printf("%d\n",f[m]);
        }
        return 0;
}