Charm Bracelet

Charm Bracelet

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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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 W_i (1 ≤ W_i ≤ 400), a 'desirability' factor D_i (1 ≤ D_i ≤ 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: W_i and D_i

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<algorithm>
#include<string.h>
using namespace std;
int dp[20000] ;
int main() {
   // freopen("a.txt" , "r" , stdin ) ;
    int w[4000] , D[4000] ;
    int M ;
    int n ;
    
    int result = 0 ;
    while ( scanf("%d%d" , &n , &M ) != EOF ) {
        for (int i = 0 ; i < n ; i++ ) {
            scanf("%d%d" , &w[i] , &D[i] ) ;
        }
        memset (dp , 0 , sizeof(dp) ) ;
        for (int i = 0 ; i < n ; i++ ) {
            for (int j = M ; j >= w[i] ; j-- ) {
                dp[j] = max ( dp[j] , dp[j - w[i]] + D[i] ) ;
            }
        }
        result = -9999999 ;
        for (int i = 0 ; i <= M ; i++ ) {
            if ( result < dp[i] )
                result = dp[i] ;
          //  printf ("%d " , dp[i] ) ;
        }

        printf("%d
" , result ) ;
    }
    return 0 ;
}