zoukankan      html  css  js  c++  java
  • POJ2782:Bin Packing

    Description

    Download as PDF
     

    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 li$ le$l<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> , ..., ln<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>(1$ le$n$ le$105)<tex2html_verbatim_mark> . The second line contains one integer that corresponds to the bin length l$ le$10000<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.

    epsfbox{p3503.eps}
     
    将所有物品由大到小排序,从大到小装入物品,如果某个装了一个物品的BIN的余量可以再装入一个物品,就把物品转入,否则就申请一个新的BIN.
     
    #include<cstdio>
    #include<algorithm>
    #include<iostream>
    using namespace std;
    int x[100002];
    int main(){
        int t;cin>>t;
        int g=1;
        while(t--){
            if(g++!=1)puts("");
            int n;cin>>n;int m;cin>>m;
            int sum=0;
            for(int i=0;i<n;i++){
                scanf("%d",x+i);
            }
            sort(x,x+n);
            int i=0,j=n-1;
            while(i<j){
                if(x[i]+x[j]<=m){
                    j--;i++;sum++;
                }else{j--;sum++;}
            }
            if(i==j)sum++;
            cout<<sum<<endl;
        }
    return 0;
    }
    View Code
  • 相关阅读:
    19.将写好的输出到本地 文件格式:Step
    18.对Topo进行打孔
    17.球体
    16.圆柱
    15.绘制圆锥
    14.Chamfer把正方体所有的边倒角
    13.绘制一个方体
    ①②坐标点
    esp8266接线
    IP解析
  • 原文地址:https://www.cnblogs.com/demodemo/p/4716079.html
Copyright © 2011-2022 走看看