POJ 3624 Charm Bracelet 01背包

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

#include<iostream>
#include <cstring>
#include<algorithm>
#include <cstdio>
#include <map>
#include <string>
using namespace std;
const int maxn=12880;
int f[maxn],w[maxn],d[maxn];
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
for(int i=1;i<=n;i++)
{
scanf("%d%d",&w[i],&d[i]);
}
memset(f,0,sizeof(f));
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;
}