Bin Packing

Bin Packing

题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=85904#problem/F

题目：

A set of n<tex2html_verbatim_mark> 1-dimensional items have to be packed in identical bins. All bins have exactly the same length l<tex2html_verbatim_mark> and each item i<tex2html_verbatim_mark> has length l_il<tex2html_verbatim_mark> . We look for a minimal number of bins q<tex2html_verbatim_mark> such that

• each bin contains at most 2 items,
• each item is packed in one of the q<tex2html_verbatim_mark> bins,
• the sum of the lengths of the items packed in a bin does not exceed l<tex2html_verbatim_mark> .

You are requested, given the integer values n<tex2html_verbatim_mark> , l<tex2html_verbatim_mark> , l₁<tex2html_verbatim_mark> , ..., l_n<tex2html_verbatim_mark> , to compute the optimal number of bins q<tex2html_verbatim_mark> .

Input

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

The first line of the input file contains the number of items n<tex2html_verbatim_mark>(1n10₅)<tex2html_verbatim_mark> . The second line contains one integer that corresponds to the bin length l10000<tex2html_verbatim_mark> . We then have n<tex2html_verbatim_mark> lines containing one integer value that represents the length of the items.

Output

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

For each input file, your program has to write the minimal number of bins required to pack all items.

Sample Input

1
 
10
80
70
15
30
35
10
80
20
35
10
30

Sample Output

6

Note: 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.

 

题意：

给定n个物品的长度a[i]，每个箱子的长度是l,同时要求每个箱子最多装两个物品。求至少要多少个箱子才能装下所有的物品。

分析：

将n个物品从长到短进行排序，比较最短值和最长值的和与箱子长度进行比较即可

 #include<iostream>
 #include<algorithm>
 using namespace std;
 int a[];
 int main()
 {
    int t;
    cin>>t;
     while(t--)
     {
        int n,l,i,j=;
        int count=;
        cin>>n>>l;
        for( i=;i<n;i++)
             cin>>a[i];
          sort(a,a+n);
        for(i=n-;i>=;i--)
        {
            if(i==j) {count++;break;}
             if(a[i]+a[j]<=l) j++;
               count++;
               if(i==j) break;
             }
                cout<<count<<endl;
                if(t)   cout<<endl;
     }
            return ;
 }