首页 > 其他 > 详细

POJ 3624 Charm Bracelet

时间:2016-09-01 18:16:04      阅读:108      评论:0      收藏:0      [点我收藏+]
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 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

Source

 
  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 }

 

POJ 3624 Charm Bracelet

原文:http://www.cnblogs.com/cumulonimbus/p/5830659.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!