Charm Bracelet (POJ 3624）

最新推荐文章于 2022-04-25 19:27:35 发布

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最新推荐文章于 2022-04-25 19:27:35 发布

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原文链接：http://www.cnblogs.com/xingkongyihao/p/7192909.html

本文介绍了一种使用一维动态规划解决最佳魅力手链问题的方法，旨在帮助读者理解如何在给定重量限制的情况下选择最优组合，以获得最高总吸引力评分。

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 40141		Accepted: 17439

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 weightW_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

23

用二维动态肯定不行，这就需要优化成一维的了。

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 using namespace std;
 5 const int N = 3410;
 6 int w[N],d[N];
 7 int dp[12881];
 8 int main(){
 9     int n,m;
10     cin>>n>>m;
11     for(int i = 0; i < n; i ++)scanf("%d %d",&w[i],&d[i]);
12     for(int i = 0; i < n; i ++){
13         for(int j = m; j >= w[i]; j--){
14             dp[j] = max(dp[j],dp[j-w[i]]+d[i]);
15         }
16     }
17     printf("%d\n",dp[m]);
18     return 0;
19 }