bookshelf-01背包问题

最新推荐文章于 2023-08-25 18:54:16 发布

Smile__wei

最新推荐文章于 2023-08-25 18:54:16 发布

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分类专栏： key to exercises 文章标签： 01背包问题

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本文探讨了一个有趣的编程问题：如何通过让部分牛堆叠起来达到书架的高度，使得未参与堆叠的牛的总高度最小。这实际上是一个01完全背包问题的变种，通过动态规划算法来解决。

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Desription

Farmer John recently bought another bookshelf for the cow library, but the shelf is getting filled up quite quickly, and now the only available space is at the top

FJ has N cows (1 ≤ N ≤ 20) each with some height of H_i (1 ≤ H_i ≤ 1,000,000 - these are very tall cows). The bookshelf has a height of B (1 ≤ B ≤ S, where S is the sum of the heights of all cows).

To reach the top of the bookshelf, one or more of the cows can stand on top of each other in a stack, so that their total height is the sum of each of their individual heights. This total height must be no less than the height of the bookshelf in order for the cows to reach the top.

Since a taller stack of cows than necessary can be dangerous, your job is to find the set of cows that produces a stack of the smallest height possible such that the stack can reach the bookshelf.

Input

* Line 1: Two space-separated integers: N and B

* Lines 2..N+1: Line i+1 contains a single integer: H_i

Output

* Line 1: A single integer representing the (non-negative) difference between the total height of the optimal set of cows and the height of the shelf.

Sample Input

Sample Output

John有N头牛每头高度为Hi，一个书架高为B，现将牛逐个踩在其他牛背上达到书架的高度B，求队列外剩下的牛的最小高度。

01完全背包问题，简化为一个载重为B的背包，每个物体质量为Hi，求背包外物体最小质量，即总质量和减去能装入背包的最大质量。

AC代码：

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int dp[2000002];
int main()
{
    int n,b;
    while(scanf("%d%d",&n,&b)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        int h[20+5];
        int sum=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&h[i]);
            sum+=h[i];
        }
        for(int i=1;i<=n;i++)
            for(int j=sum;j>=h[i];j--)
                dp[j]=max(dp[j],dp[j-h[i]]+h[i]);
        int m=sum;
        for(int i=b;i<=sum;i++)
            if(dp[i]>=b&&dp[i]-b<m)m=dp[i]-b;
        cout<<m<<endl;
    }
    return 0;
}