POJ 3624 Charm Bracelet
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 13771 | Accepted: 6272 |
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
1 #include<stdio.h> 2 #include<string.h> 3 int f[13000] ; 4 int main(){ 5 int n,m; 6 while(scanf("%d%d",&n,&m)!=EOF){ 7 memset(f,0,sizeof(f)) ; 8 for(int x=1;x<=n;x++){ 9 int w,d ; 10 scanf("%d%d",&w,&d) ; 11 for(int i=m;i>=w;i--) 12 if(f[i-w]+d > f[i]) 13 f[i] = f[i-w] + d; 14 } 15 printf("%d\n",f[m]) ; 16 } 17 return 0; 18 }