POJ 3624 Charm Bracelet

最新推荐文章于 2022-08-11 11:21:35 发布

baymax520

最新推荐文章于 2022-08-11 11:21:35 发布

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分类专栏：背包dp 文章标签： dp

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本文探讨了一个典型的01背包问题实例，通过分析题目要求，给出了一种有效的动态规划解决方案，并详细介绍了如何优化内存使用，确保算法在限定的内存资源内高效运行。

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Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 44162		Accepted: 18976

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

经典的01背包问题，用dp[i][j]表示在前i个物品中选择体积不超过j的最大价值。状态转移方程dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+d[i])，但是这一题中如果开二维数组肯定会超内存，所以我们可以优化内存空间，观察可知dp[i][j]的值都是由i-1时的状态推出来的，所以一维数组即可。即dp[j]=max(dp[j],dp[j-w[i]]+d[i])

还要注意因为是一维的了，所以在for循环遍历的时候j要倒序，因为顺序的话就有可能使用到之前已经更新过的状态。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
   int dp[15000],w[5000],d[5000];
   int n,m,i,j;
   while(scanf("%d %d",&n,&m)!=EOF)
   {
       for(i=1;i<=n;i++)
        scanf("%d %d",&w[i],&d[i]);
       memset(dp,0,sizeof(dp));
       for(i=1;i<=n;i++)
        for(j=m;j>=w[i];j--)
       {
           dp[j]=max(dp[j],dp[j-w[i]]+d[i]);
       }
       printf("%d\n",dp[m]);
   }
}