Title Description
Known load of a backpack and a M n of items, item number from 0 to n-1. By weight of the i-th items of wi, if the i-th item into the backpack will benefit pi, where, wi> 0, pi> 0,0 <= i <n. The so-called 0/1 knapsack problem is the inability to split the article, only the whole into a bag or case is not loaded, loading seeking a preferred embodiment such that the maximum total revenue.
Note:
1, please solve this problem by backtracking (to use bounding function prune).
2, all the data have been tested by pi / wi descending order.
Entry
Line 1 has two positive integers n (n <= 50) and M, expressed n items, the backpack load of M (m <= 100). Then enter the weight of the item number n, the n value of the final input return objects.
Export
The total revenue of the optimal loading plan
c ++ code is as follows:
#include<iostream>
using
namespace
std;
int
n;
int
m;
int
x[100];
int
y[100];
int
fp=0;
int
Bound(
int
k,
int
cp,
int
cw,
int
*w,
int
*p)
{
int
b=cp,c=cw;
for
(
int
i=k+1;i<n;i++)
{
c+=w[i];
if
(c<m)
b+=p[i];
else
return
(b+(1-(c-m)/w[i])*p[i]);
}
return
b;
}
void
BK(
int
k,
int
cp,
int
cw,
int
&fp,
int
*x,
int
*y,
int
*w,
int
*p)
{
int
j;
int
bp;
if
(cw+w[k]<=m)
{
y[k]=1;
if
(k<n-1) BK(k+1,cp+p[k],cw+w[k],fp,x,y,w,p);
if
(cp+p[k]>fp&&k==n-1)
{
fp=cp+p[k];
for
(j=0;j<=k;j++)
x[j]=y[j];
}
}
if
(Bound(k,cp,cw,w,p)>=fp)
{
y[k]=0;
if
(k<n-1)
BK(k+1,cp,cw,fp,x,y,w,p);
if
(cp>fp&&k==n-1)
{
fp=cp;
for
(j=0;j<=k;j++)
x[j]=y[j];
}
}
}
int
BK(
int
*x,
int
*w,
int
*p)
{
int
y[100]={0};
int
fp;
BK(0,0,0,fp,x,y,w,p);
return
fp;
}
int
main()
{
int
i;
int
w[100];
int
p[100];
cin>>n>>m;
for
(i=0;i<n;i++)
cin>>w[i];
for
(i=0;i<n;i++)
cin>>p[i];
cout<<BK(x,w,p);
return
0;
}
