Charm Bracelet

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

Submit Status

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

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

01背包

 

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