poj 3628 Bookshelf 2（0/1背包）

最新推荐文章于 2019-07-15 10:37:49 发布

Misdom_Tian_Ya

最新推荐文章于 2019-07-15 10:37:49 发布

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分类专栏：动态规划——背包动态规划

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动态规划——背包

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本文介绍了一种通过0-1背包算法解决最小堆叠高度问题的方法。问题要求找到一组奶牛，使它们的堆叠高度刚好达到或超过书架顶部的高度，同时这个高度又尽可能接近书架高度，以确保安全。

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Description

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. Your program should print the minimal 'excess' height between the optimal stack of cows and 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

描述：1、总高度不应小于书架高度。
             2、求最小高度。
             3、输出的是最小高度与书架高度的差。
此题利用0-1背包求，求出所有的取值可能，然后再枚举求解。

#include <iostream>
#include <math.h>
using namespace std;
int H[21],dp[10000000];
int main()
{
    int N,B,i,j,sum=0;
    cin>>N>>B;
    for(i=0;i<N;i++)
    {
        cin>>H[i];
        sum=sum+H[i];
    }
    for(i=0;i<=sum;i++)
        dp[i]=0;
    for(i=0;i<N;i++)
        for(j=sum;j>=H[i];j--)
          dp[j]=max(dp[j-H[i]]+H[i],dp[j]);
    int min1=100000000;
    for(i=0;i<=sum;i++)
    {
        if(dp[i]>=B)
            min1=min(min1,dp[i]);
    }
    cout<<min1-B<<endl;
    return 0;
}