Charm Bracelet
Time Limit: 1000MS | | Memory Limit: 65536K |
Total Submissions: 24165 | | Accepted: 10898 |
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 weightWi (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<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
int i,j,n,m;
int a[3500],b[3500],dp[13000];
while(scanf("%d %d",&n,&m)!=EOF)
{
for(i=1;i<=n;i++)
{
scanf("%d %d",&a[i],&b[i]);
}
memset(dp,0,sizeof(dp));
for(i=1;i<=n;i++)
{
for(j=m;j>=a[i];j--)
dp[j]=max(dp[j],dp[j-a[i]]+b[i]);
}
printf("%d\n",dp[m]);
}
return 0;
}