poj-3624Charm Bracelet

谁践踏了优雅 2022-08-09 13:39 2阅读 0赞

Charm Bracelet

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;28601</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;12817</td> 
  </tr> 
 </tbody> 
</table>

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

简单01背包：

#include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    int f[13000],val[3500],vol[3500];
    int main()
    {
    	int m,n,i,j;
    	scanf("%d%d",&m,&n);
    	for(i=0;i<m;i++)
    		scanf("%d%d",&vol[i],&val[i]);
    		memset(f,0,sizeof(f));
    	for(i=0;i<m;i++)
    	{
    		for(j=n;j>=vol[i];j--)
    		{
    			f[j]=max(f[j],f[j-vol[i]]+val[i]);
    		}
    	}
    	printf("%d\n",f[n]);
    	return 0;
    }