poj-3624Charm Bracelet

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 28601		Accepted: 12817

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

简单01背包：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int f[13000],val[3500],vol[3500];
int main()
{
	int m,n,i,j;
	scanf("%d%d",&m,&n);
	for(i=0;i<m;i++)
		scanf("%d%d",&vol[i],&val[i]);
		memset(f,0,sizeof(f));
	for(i=0;i<m;i++)
	{
		for(j=n;j>=vol[i];j--)
		{
			f[j]=max(f[j],f[j-vol[i]]+val[i]);
		}
	}
	printf("%d\n",f[n]);
	return 0;
}