贪心算法之木材装集装箱问题

                                   贪心算法应用

                                             ——木材的装箱问题

一、问题描述：

给定一个集装箱，其长为L、宽为W、高为H。现有一批圆柱型木材，每根木材的长均为L，但是半径r不同。设第i根木材半径为ri。

问：如何装箱，使得集装箱的空间利用率最高？

 

 

 

 

二、算法设计：

——采用贪心算法思想，将箱子抽象为L*W*H的长方体，将木材抽象为底面半径为ri、长为L的圆柱体。

算法步骤：

  1、首先对圆柱体按半径大小从大到小进行排序

  2、取出圆柱体中底面半径最大的一个，从左下角坐标开始检查矩形空闲位置并判断当前圆柱体是否可以放入（判断圆柱体底面圆的圆心距是否合适，以及底面面积是否超过了空闲矩形的边框）。

  3、接下来再将剩余的圆柱体取出，重复步骤2直至矩形空间中不再能够容纳下剩余圆柱体中底

三、程序实现：

#include <iostream>#include <stdio.h>#include <graphics.h>#include <conio.h>#include <math.h>//#include <stdlib.h>using namespace std;#define MAXNUM 100#define pi 3.141592653589int columnum,count;float H,W;struct Circle_wood //圆柱形木头的数据结构{double x,y,r;int sel;      //判断是否和别人相交}cr[MAXNUM];double dis(Circle_wood a,double x,double y) //两个圆圆心距{return sqrt((a.x-x)*(a.x-x)+(a.y-y)*(a.y-y));}int sort(Circle_wood *wood,int n)//按圆柱半径由大到小排序  {Circle_wood a;int j;for(int i=0;i<columnum;i++){ a=wood[i];j=i-1;while(j>=0 && wood[j].r<a.r){wood[j+1]=wood[j];j=j-1;}wood[j+1]=a;}return 0;}int put(int now,double tx,double ty)//判断在位置(tx,ty)是否可以放下{if(tx<cr[now].r || tx+cr[now].r>W || ty<cr[now].r || ty+cr[now].r>H)return 0; //不能放下，越界for(int i=0;i<now;i++)if(cr[i].sel)//判断是否和别人相交if(dis(cr[i],tx,ty)<(cr[i].r+cr[now].r-0.01)) //相交return 0;return 1;}void check(int now)//为第now个木头找位置{ double t,dx,dy;for(int i=0;i<now;i++)if(cr[i].sel)//对每个已插入的木头{t=cr[i].r+cr[now].r;for(int j=0;j<=360; j++)//对每个角度进行检查{ dx=cr[i].x+t*cos(-pi/2.0 + j*pi/180.0);dy=cr[i].y+t*sin(-pi/2.0 + j*pi/180.0);if(put(now,dx,dy)==1)//找到合适位置了{ count++;cr[now].x=dx;cr[now].y=dy;cr[now].sel=1;return ;}}}}void Gdy()//确定坐标并实现木材的定位{count=0;int i;for(i=0;i<columnum;i++) //初始化cr[i].sel=0; for(i=0;i<columnum;i++){if(count==0 && cr[i].r <= H/2 &&cr[i].r <=W/2)//第一个能放进的{cr[i].x=cr[i].r;cr[i].y=cr[i].r;cr[i].sel=1;count++;}elsecheck(i); //检查其他木头}}void draw()//画图{int t=0;int k,j;char *S[MAXNUM]={"0 ","1 ","2 ","3 ","4 ","5 ","6 ","7 ","8 ","9 ","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26","27","28","29","30","31","32","33","34","35","36","37","38","39","40","41","42","43","44","45"};/*int *S[20];for(k=0,j=0;k<columnum;k++){if(cr[k].sel==1)        {    *S[j]=(int)cr[k].r;   j++;}}*/HBRUSH hbrush_old;HDC    hdc     = GetWindowDC( GetDesktopWindow() );HPEN   hpen1   = CreatePen( PS_SOLID, 1, RGB(255,200,200) );// 创建浅色1像素宽度的实线画笔 HPEN   hpen2   = CreatePen( PS_DASH, 5, RGB(0,255,0) ); // 创建绿色5像素宽度的破折画笔HBRUSH hbrush1 = CreateSolidBrush( RGB(60,60,60) ); // 创建一个实体蓝色画刷 HBRUSH hbrush2 = CreateSolidBrush( RGB(255,255,255) ); //紫色 //HBRUSH hbrush3 = CreateSolidBrush( RGB(0,255,64) );//绿色//HBRUSH hbrush4 = CreateSolidBrush( RGB(128,255,0) );//HBRUSH hbrush5 = CreateSolidBrush( RGB(255,0,128) );//紫色HPEN   hpen_old= (HPEN)SelectObject( hdc, hpen1 );hbrush_old     = (HBRUSH)SelectObject( hdc, hbrush1);Rectangle(hdc,200,400,200+(int)(W*20),400-(int)(H*20)); for(int i=0;i<columnum;i++){if(i%4==0)hbrush_old = (HBRUSH)SelectObject( hdc, hbrush2);else if(i%4==1)hbrush_old = (HBRUSH)SelectObject( hdc, hbrush2);else if(i%4==2) hbrush_old = (HBRUSH)SelectObject( hdc, hbrush2);elsehbrush_old = (HBRUSH)SelectObject( hdc, hbrush2);if(cr[i].sel){t=i;Ellipse( hdc,(int)(200+20*(cr[i].x-cr[i].r)), (int)(400-20*(cr[i].y+cr[i].r)), (int)(200+20*(cr[i].x+cr[i].r)), (int)(400-20*(cr[i].y-cr[i].r)) );//画圆TextOut( hdc,(int)(200+20*cr[i].x),(int)(400-20*cr[i].y),S[i+1],2); //i,signed i+1; 并用数字标记//TextOut( hdc,(int)(200+20*cr[i].x),(int)(400-20*cr[i].y),cr[i].r,2); //i,signed i+1; 并用数字标记}Sleep(750);}TextOut( hdc,(int)(200+20*cr[t].x),(int)(400-20*cr[t].y),S[t+1],2); //最后一个圆柱形木头放入}int main()//主函数{FILE *fp;char filename[20];long start; //用以计算程序运行时间double area =0;int i;//printf(" 请输入数据文件:"); //读入圆柱形木头数据//scanf("%s",filename);if((fp=fopen("123.txt","r"))==NULL){printf(" 输入文件不存在!\n ");//文件不存在则退出exit(0);}start=GetTickCount();//fscanf(fp,"%d",&columnum);cout<<"请输入圆柱个数：";cin>>columnum;cout<<"集装箱的宽：";cin>>W;cout<<"集装箱的高：";cin>>H;//cout<<"集装箱的长：";//cin>>L;for(i=0;i<columnum;i++) fscanf(fp,"%lf",&cr[i].r); sort(cr,columnum); //先进行对木头的排序Gdy(); //这个是本程序的主体函数for(i=0;i<columnum;i++) //计算一个木头所占有的面积if(cr[i].sel) area += pi*cr[i].r*cr[i].r; draw(); //画出图形//printf(" %ld\n",columnum);//printf("%lf\n%lf\n%lf\n",W,H);for(i=0;i<columnum && cr[i].sel==1;i++){        cout<<"可装入的圆柱序号为"<<i<<endl;}printf(" 装入货物的面积为:%lf\n",area);printf(" 集装箱利用率为:%lf%%\n",area/(W*H)*100);//计算集装箱利用率printf(" 程序运行时间为:%f秒\n ",(GetTickCount()-start)/1000.0); //计算程序运行时间getchar();getchar();return 0;}

四、程序结果：

 

五、算法效率

考虑的最坏情况，每个圆定位都要遍历之前n-1个圆，角度要经过360次计算，要经过n-1次计算与其他圆是否相交，故最差为O(n^3)，最好是O(n^2)