（简单贪心）Bin Packing

最新推荐文章于 2025-09-09 12:24:46 发布

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本文探讨了一种特定条件下的物品装箱问题：在一维空间内将不同长度的物品放入固定大小的箱子中，每个箱子最多只能放置两个物品且总长度不超过箱子的容量。文章提出了一种基于贪心算法的解决方案，并给出了示例输入输出及实现代码。

A set of n 1-dimensional items have to be packed in identical bins. All bins have exactly the same length l and each item i has length li<=l . We look for a minimal number of bins q such that

each bin contains at most 2 items,
each item is packed in one of the q bins,
the sum of the lengths of the items packed in a bin does not exceed l .

You are requested, given the integer values n , l , l1 , ..., ln , to compute the optimal number of bins q .

Input

The first line of the input contains the number of items n (1<=n<=10 ⁵) . The second line contains one integer that corresponds to the bin length l<=10000 . We then have n lines containing one integer value that represents the length of the items.

Output

Your program has to write the minimal number of bins required to pack all items.

Sample Input

Sample Output

Hint

The sample instance and an optimal solution is shown in the figure below. Items are numbered from 1 to 10 according to the input order.

思路：贪心

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
#include<iostream>
int n,l;
int a[100010];
int main()
{
        cin>>n>>l;
        for(int i=0;i<n;++i)
            cin>>a[i];
        sort(a,a+n);
        int sum=0;
        int len=0;
        int temp=n-1;
        for(int i=0;i<n;i++)
         if(a[i]>0)
         {
             if(a[i]<l)
                {
                    for(int j=temp;j>i;--j)
                    {
                        if(a[j]>0&&a[j]+a[i]<=l)
                            {
                                a[j]=-1;
                                temp--;
                                break;
                            }
                    }
                }
            a[i]=-1;
            sum++;
         }
        printf("%d\n",sum);
}