| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 10399 | Accepted: 4667 |
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
Source
#include<cstdio>
#include<iostream>
using namespace std;
const int mm=13333;
int f[mm],w[mm],d[mm];
int i,j,k,n,m;
int main()
{
while(scanf("%d%d",&n,&m)!=-1)
{
for(i=0;i<n;++i)scanf("%d%d",&w[i],&d[i]);
for(i=0;i<m;++i)f[i]=0;
for(i=0;i<n;++i)
for(j=m;j>=w[i];--j)f[j]=max(f[j],f[j-w[i]]+d[i]);
printf("%d\n",f[m]);
}
return 0;
}