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; }