B. Shaass and Bookshelf
Shaass has n books. He wants to make a bookshelf for all his books. He wants the bookshelf's dimensions to be as small as possible. The thickness of the i-th book is ti and its pages' width is equal to wi. The thickness of each book is either 1 or 2. All books have the same page heights.
Shaass puts the books on the bookshelf in the following way. First he selects some of the books and put them vertically. Then he puts the rest of the books horizontally above the vertical books. The sum of the widths of the horizontal books must be no more than the total thickness of the vertical books. A sample arrangement of the books is depicted in the figure.
Help Shaass to find the minimum total thickness of the vertical books that we can achieve.
Input
The first line of the input contains an integer n, (1 ≤ n ≤ 100). Each of the next n lines contains two integers ti and wi denoting the thickness and width of the i-th book correspondingly, (1 ≤ ti ≤ 2, 1 ≤ wi ≤ 100).
Output
On the only line of the output print the minimum total thickness of the vertical books that we can achieve.
Sample test(s)
input
5 1 12 1 3 2 15 2 5 2 1
output
5
input
3 1 10 2 1 2 4
output
3
dp code but unable to decode decode latter
#include <bits/stdc++.h>
using namespace std;
int w[100],t[100],dp[303];
int main(){
int n,i,j,sum=0;cin>>n;
for(i=0;i<n;++i)
cin>>t[i]>>w[i],sum+=t[i];
for(i=0;i<n;++i)
for(j=sum;j>=t[i]+w[i];--j)
dp[j]= max(dp[j],dp[j-t[i]-w[i]]+t[i]);
cout<<sum-dp[sum]<<endl;
}
-------------------------------------------------gready-----------------------------------------------------------------
As said in the statement, the thickness of each book is either 1 or 2. Think about when we want to arrange v1 books of thickness 1 andv2 books of thickness 2 vertically and arrange all other n - v1 - v2 books horizontally above them to achieve a configuration with total thickness of vertical books equal to v1 + 2v2. Is it possible to find such arrangement? Because the total thickness of vertical books is fixed it's good to calculate the minimum possible total width of horizontal books. As the width of a book doesn't matter in vertical arrangement it's good to use the books with shorter width horizontally and the ones with longer width vertically. So pick out v1 books with longest width among books of thickness 1 and do the same with books of thickness 2. The sum of width of n - v1 - v2 remaining books should be at most v1 + 2v2.
The solution would be to try the things we explained above for all possible values of v1 and v2. And print the best answer. :)
There exists other ways to solve this problem mostly using dynamic programming but this was the intended solution of the problem.
Here is a nice implementation in C++ from contestant Bayram: 3485189 (You should also know that 'bir' means 'one' in Turkish and 'iki' means two!)
now think dp approach since if width will not be 1 , 2 that it will be quit tough by this approach
-----------------------------------------------------code---------------------------------------------------------------
#include<bits/stdc++.h>
using namespace std;
int main()
{
int ans=0;
int n;
cin>>n;
int p=0,q=0;
int sum=0;
int o[109],t[109];
for(int i=0;i<n;i++)
{
int a,b;
cin>>a>>b;
if(a==1) o[p++]=b;
else t[q++]=b;
sum+=b;
}
sort(o,o+p);
sort(t,t+q);
//for(int i=0;i<p;i++)cout<<o[i]<<" ";cout<<endl;
// for(int j=0;j<q;j++) cout<<t[j]<<" ";
int st1=p-1;
int st2=q-1;
int base=0;
while(st1>=0 && st2>=0)
{
// cout<<"sts "<<st1<<" "<<st2<<"sum "<<sum<<endl;
if(sum-o[st1]<=base+1)
{
cout<<base+1<<endl;
return 0;
}
else if(st1>=1 && sum-o[st1]-o[st1-1]<=base+2)
{
cout<<base+2<<endl;
return 0;
}
else if(sum-t[st2]<=base+2)
{
cout<<base+2<<endl;
return 0;
}
else if(sum-t[st2]-o[st1]<=base+3)
{
cout<<base+3<<endl;
return 0;
}
else
{
int bb=0;
// cout<<st1<<" "<<st2<<endl;
int max1=o[st1];
bb=1;
// cout<<"a "<<max1<<endl;
if(st1>=1)
{
max1+=o[st1-1];bb++;
}
//cout<<"b "<<max1<<endl;
int max2=t[st2];
// cout<<"maxi1 "<<max1<<" max2 "<<max2<<endl;
if(max1>max2)
{
base+=bb;
st1-=2;
sum-=max1;
// base+=2;
}
else
{
st2-=1;
sum-=max2;
base+=2;
}
}
// cout<<"sts "<<st1<<" "<<st2<<" base "<<base<<" sum "<<sum<<endl;
}
if(st1<0)
{
while(st2>=0)
{
sum-=t[st2];
base+=2;
if(base>=sum)
{
cout<<base<<endl;
return 0;
}
st2-=1;
}
}
if(st2<0)
{
while(st1>=0)
{
sum-=o[st1];
base+=1;
if(base>=sum)
{
cout<<base<<endl;
return 0;
}
st2-=1;
}
}
}
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