The Theatre Problem
In this problem the basic idea is to make all combination that are possible in this.
To get maximum amount of money through movies you have you have to pickup maximum demand of particular movies and doesn’t clash with maximum demand of other movie in that time zone.
It can be easily understand just by looking on code:
/*The Theatre Problem*/
/****killing like kamikaze****/
/*tanuj yadav*/
/**********************/
/*the threatre problrm*/
#include<bits/stdc++.h>
using namespace std;
int main()
{
int t;
cin>>t;
long int ans=0;
for(int i=0;i<t;i++)
{
int n;
cin>>n;
char s;
int y,x;
int a[4][4];
for(int j=0;j<4;j++)
{
for(int k=0;k<4;k++)
{a[j][k] = 0;}
}
for(int j=0;j<n;j++)
{
cin>>s>>y;
x = int(s)-65;
if(y==12)
{y=0;}
else if(y==3)
{y=1;}
else if(y==6)
{y=2;}
else if(y==9)
{y=3;}
a[x][y]+=1;
}
int countin =0;
int visit[4] = {0};
long int sum;
long int mx = -10000000;
for(int j=0;j<4;j++)
{sum=0;
countin =0;
int nw[4] ={0};
visit[j] = 1;
for(int k=0;k<4;k++)
{
if(visit[k]!=1)
{visit[k]=1;
for(int l=0;l<4;l++)
{
if(visit[l]!=1)
{visit[l]=1;
for(int m=0;m<4;m++)
{
if(visit[m]!=1)
{
nw[0] = a[0][j];
nw[1] = a[1][k];
nw[2] = a[2][l];
nw[3] = a[3][m];
for(int w=0;w<4;w++)
{if(nw[w]==0)
{countin++;}
}
sort(nw,nw+4,greater<int>());
sum = 100*nw[0]+75*nw[1]+50*nw[2]+25*nw[3];
sum = sum -100*countin;
if(sum>mx)
{mx=sum;}
countin=0;
}
}
visit[l]=0;
}
}
visit[k] = 0;
} }
visit[j] = 0;}
cout<<mx<<endl;
ans = ans + mx; }
cout<<ans;
return 0;
}