POJ 3624 Charm Bracelet(01背包)

最新推荐文章于 2019-08-06 15:11:09 发布

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原文链接：http://www.cnblogs.com/GO-NO-1/p/5974728.html

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#c/c++

本文介绍了一种典型的01背包问题——魅力手链问题。该问题要求在重量限制下选择最优组合的装饰物以获得最高的魅力值总和。文章提供了一个C++实现的例子，展示了如何通过动态规划解决这个问题。

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 34532		Accepted: 15301

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

Sample Output

Source

USACO 2007 December Silver

感觉数据有点水，典型01背包

 1 //Accepted    212K    266MS    C++    395B
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<string.h>
 5 using namespace std;
 6 
 7 int dp[13000];
 8 int w[3500], d[3500];
 9 int main(void)
10 {
11     int n,m;
12     while(scanf("%d%d", &n,&m)!=EOF){
13         for(int i=0;i<n;i++)
14             scanf("%d%d",&w[i],&d[i]);
15         memset(dp, 0 ,sizeof(dp));
16         for(int i=0;i<n;i++)
17             for(int j=m;j>=w[i];j--)
18                 dp[j] = max(dp[j], dp[j-w[i]] + d[i]);
19         printf("%d\n", dp[m]);
20     }
21     return 0;
22 }