## POJ - 3624 Charm Bracelet

jlqwer 发表于 代码 分类，标签: 热度:91°
• 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 <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main()
{
int N,M,W[3500],D[3500],dp[3500];
while(cin>>N>>M)
{
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-W[i]]+D[i],dp[j]);
}
}
cout<<dp[M]<<endl;
}
return 0;
}```

lll