poj 3624 Charm Bracelet

http://poj.org/problem?id=3624

一维数组的01背包，v一定要逆序枚举，不然会多次放同一种物品。

                                                                                  Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 24456		Accepted: 11031

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 W_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 6
1 4
2 6
3 12
2 7

Sample Output

23

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int c,n;
int v[3412],w[3412],f[12888];
void backpack()
{
    memset(f,0,sizeof(f));
    int i,j;
    for(i=1;i<=n;i++)
      for(j=c;j>=0;j--)
         if(j>=v[i])
            f[j]=max(f[j],f[j-v[i]]+w[i]);
     printf("%d\n",f[c]);

}
int main()
{
    int i;
    scanf("%d%d",&n,&c);
    for(i=1;i<=n;i++)
      scanf("%d%d",&v[i],&w[i]);
    backpack();
    return 0;

}