POJ 3624 Charm Bracelet

Posted 2020-08-05 Cumulonimbus

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 34193		Accepted: 15154

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 W_i (1 ≤ W_i ≤ 400), a ‘desirability‘ factor D_i (1 ≤ D_i ≤ 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: W_i and D_i

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

Source

USACO 2007 December Silver

 

  01背包问题。

 

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstring>
 4 using namespace std;
 5 
 6 const int MAX_N=3500;
 7 const int MAX_M=13000;
 8 
 9 int v[MAX_N],w[MAX_N];
10 int f[MAX_M];
11 int main()
12 {
13     int N,M,V,W;
14     while(scanf("%d%d",&N,&M)==2){
15         memset(f,0,sizeof(f));
16         for(int i=1;i<=N;i++){
17             scanf("%d%d",&V,&W);
18             for(int j=M;j>=0;j--)
19                 if(j>=V) f[j]=max(f[j], f[j-V]+W);
20         }
21         printf("%d\n",f[M]);
22     }
23 }