POJ 3624 Charm Bracelet

最新推荐文章于 2019-08-06 15:11:09 发布

wintermelon27

最新推荐文章于 2019-08-06 15:11:09 发布

阅读量800

点赞数

CC 4.0 BY-SA版权

分类专栏： Dynamic Programming 文章标签：背包

本文链接：https://blog.youkuaiyun.com/ACMWONDER/article/details/24442057

Dynamic Programming 专栏收录该内容

6 篇文章

订阅专栏

本文介绍了一个经典的01背包问题案例，通过分析Bessie选择最佳魅力装饰品的问题，讲解了如何在给定重量限制下获得最大价值总和的算法实现过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 20499		Accepted: 9293

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

Source

USACO 2007 December Silver

题解：经典背包问题，基础01背包，状态转移方程：

f[v]=max(f[v],f[v-a[i].w]+a[i].d);

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int n,m;
int f[15000];
struct thing
{
  int w;
  int d;
}a[5000];
int main()
{
    //freopen("in.txt","r",stdin);
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
      scanf("%d%d",&a[i].w,&a[i].d);
      f[i]=0;
    }
    for(int i=1;i<=n;i++)
    {
      for(int v=m;v>=a[i].w;v--)
      {
        f[v]=max(f[v],f[v-a[i].w]+a[i].d);
      }
    }
    //for(int i=1;i<=m;i++)printf("%d\n",f[i]);
    printf("%d\n",f[m]);
    return 0;
}