POJ-3624 Charm Bracelet（动态规划）

POJ-3624 Charm Bracelet（动态规划）

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<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define MAX 4000
int dp[MAX],w[MAX],v[MAX];
int main()
{
     int i,j,n,m;
     memset(dp,0,sizeof(dp));
       scanf("%d %d",&n,&m);
 
                             //下标从1开始与实际结合
     for(i=1;i<=n;i++)
     scanf("%d%d",&w[i],&v[i]);
     for(i=1;i<=n;i++)
     for(j=m;j>=w[i];j--)
     {
      dp[j]=max(dp[j],dp[j-w[i]]+v[i]);  //dp[j]记录最大价值
     }
     printf("%d\n",dp[m]);
     return 0;
}