POJ 3624 （01背包一维求解）

左手的ㄟ右手 2024-02-17 19:31 37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;36983</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;16186</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

AC代码：

#include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    const int max1=3402+2;
    const int max2=12880+10;
    int dp[max2];
    int w[max1],d[max1];
    int main(){
    	int n,m;
    	while(scanf("%d%d",&n,&m)==2){
    		for(int i=1;i<=n;i++){
    			scanf("%d%d",&w[i],&d[i]);
    		}
    		memset(dp,0,sizeof(dp));
    		for(int i=1;i<=n;i++){
    			for(int j=m;j>=0;j--){
    				if(j>=w[i])dp[j]=max(dp[j],dp[j-w[i]]+d[i]);
    			}
    		}
    		printf("%d\n",dp[m]);
    	}
    	return 0;
    }