Charm Bracelet
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 38909 | Accepted: 16862 |
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 <iostream>
#include <algorithm>
#include <stdio.h>
using namespace std;
int w[],d[],dp[];
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
for(int i=;i<=n;i++)
scanf("%d%d",&w[i],&d[i]);
for(int i=;i<=n;i++)
{
for(int j=m;j>=w[i];j--)
{
dp[j]=max(dp[j],dp[j-w[i]]+d[i]);
}
}
printf("%d\n",dp[m]);
}
return ;
}