Charm Bracelet（POJ-3624）

最新推荐文章于 2022-03-25 21:17:10 发布

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最新推荐文章于 2022-03-25 21:17:10 发布

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分类专栏： 01背包文章标签： acm 背包

本文链接：https://blog.youkuaiyun.com/qq_36949416/article/details/72628722

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本文介绍了一种解决最优魅力手链问题的动态规划算法。该问题要求在重量限制下选择一系列具有不同权重和吸引力的装饰品，使得总吸引力最大。文章详细解释了输入输出格式，并提供了实现代码。

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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

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int main()
{
    int n,m;
    int a[5005],b[5005],dp[12885];
    scanf("%d%d",&n,&m);
    for(int i=0; i<n; i++)
    {
        scanf("%d%d",&a[i],&b[i]);
    }
    for(int i=0; i<=m; i++)
    {
        dp[i]=0;
    }
    for(int i=0; i<n; i++)
    {
        for(int j=m; j>=a[i]; j--)
        {
            dp[j]=max(dp[j],dp[j-a[i]]+b[i]);
        }
    }
    printf("%d\n",dp[m]);
}