POJ 3624 Charm Bracelet(01背包裸题)

Charm Bracelet

Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 38909   Accepted: 16862

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

题目链接:http://poj.org/problem?id=3624
分析:01背包裸题,顺带敲了一遍,做个复习吧,怕忘了!详解请参看我的博客/ECJTUACM-873284962/p/6815610.html,里面有对01背包的完全介绍以及用法!
下面给出AC代码:
 1 #include <iostream>
 2 #include <algorithm>
 3 #include <stdio.h>
 4 using namespace std;
 5 int w[35000],d[35000],dp[35000];
 6 int main()
 7 {
 8     int n,m;
 9     while(scanf("%d%d",&n,&m)!=EOF)
10     {
11         for(int i=1;i<=n;i++)
12             scanf("%d%d",&w[i],&d[i]);
13         for(int i=1;i<=n;i++)
14         {
15             for(int j=m;j>=w[i];j--)
16             {
17                 dp[j]=max(dp[j],dp[j-w[i]]+d[i]);
18             }
19         }
20         printf("%d\n",dp[m]);
21     }
22     return 0;
23 }

 

posted @ 2017-05-08 22:50 Angel_Kitty 阅读(...) 评论(...) 编辑 收藏

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