Ecological Bin Packing
|
Background
Bin
packing, or the placement of objects of certain weights into different bins
subject to certain constraints, is an historically interesting problem. Some
bin packing problems are NP-complete but are amenable to dynamic programming
solutions or to approximately optimal heuristic solutions.
In
this problem you will be solving a bin packing problem that deals with
recycling glass.
The Problem
Recycling
glass requires that the glass be separated by color into one of three
categories: brown glass, green glass, and clear glass. In this problem you will
be given three recycling bins, each containing a specified number of brown,
green and clear bottles. In order to be recycled, the bottles will need to be
moved so that each bin contains bottles of only one color.
The
problem is to minimize the number of bottles that are moved. You may assume
that the only problem is to minimize the number of movements between boxes.
For
the purposes of this problem, each bin has infinite capacity and the only
constraint is moving the bottles so that each bin contains bottles of a single
color. The total number of bottles will never exceed 2^31.
The Input
The
input consists of a series of lines with each line containing 9 integers. The
first three integers on a line represent the number of brown, green, and clear
bottles (respectively) in bin number 1, the second three represent the number
of brown, green and clear bottles (respectively) in bin number 2, and the last
three integers represent the number of brown, green, and clear bottles
(respectively) in bin number 3. For example, the line 10 15 20 30 12 8 15 8 31
indicates
that there are 20 clear bottles in bin 1, 12 green bottles in bin 2, and 15
brown bottles in bin 3.
Integers
on a line will be separated by one or more spaces. Your program should process
all lines in the input file.
The Output
For
each line of input there will be one line of output indicating what color
bottles go in what bin to minimize the number of bottle movements. You should
also print the minimum number of bottle movements.
The
output should consist of a string of the three upper case characters 'G', 'B',
'C' (representing the colors green, brown, and clear) representing the color
associated with each bin.
The
first character of the string represents the color associated with the first
bin, the second character of the string represents the color associated with
the second bin, and the third character represents the color associated with
the third bin.
The
integer indicating the minimum number of bottle movements should follow the
string.
If
more than one order of brown, green, and clear bins yields the minimum number
of movements then the alphabetically first string representing a minimal
configuration should be printed.
Sample Input
1 2 3 4 5 6 7 8 9
5 10 5 20 10 5 10 20 10
Sample Output
BCG 30
CBG 50
Solution :
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
static int b[3],g[3],c[3];
#define max(a,b,c)
(a>b&&a>c)?a:(b>a&&b>c)?b:c
int* change(char C)
{
return (C=='B')?b:(C=='G')?g:c;
}
int main()
{
int i,min,minI,j,cur,k;
char a[6][4] = {"BCG" ,"BGC"
,"CBG" ,"CGB" ,"GBC" ,"GCB"};
while(scanf("%d %d %d %d %d %d %d %d
%d",b,g,c,b+1,g+1,c+1,b+2,g+2,c+2)==9)
{
min = 1<<30;
minI = -1;
for(i=0;i<6;i++)
{
cur =0;
for(j=0;j<3;j++)
{
for(k=0;k<3;k++)
if(k!=j)
cur+=change(a[i][j])[k];
}
if(cur<min)
{
min = cur;
minI = i;
}
}
printf("%s
%d\n",a[minI],min);
}
return 0;
}
No comments:
Post a Comment