| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 34532 | Accepted: 15301 |
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 //Accepted 212K 266MS C++ 395B 2 #include<iostream> 3 #include<algorithm> 4 #include<string.h> 5 using namespace std; 6 7 int dp[13000]; 8 int w[3500], d[3500]; 9 int main(void) 10 { 11 int n,m; 12 while(scanf("%d%d", &n,&m)!=EOF){ 13 for(int i=0;i<n;i++) 14 scanf("%d%d",&w[i],&d[i]); 15 memset(dp, 0 ,sizeof(dp)); 16 for(int i=0;i<n;i++) 17 for(int j=m;j>=w[i];j--) 18 dp[j] = max(dp[j], dp[j-w[i]] + d[i]); 19 printf("%d ", dp[m]); 20 } 21 return 0; 22 }