N meetings in one room.cpp

PROBLEM:

There is one meeting room in a firm. There are N meetings in the form of (S[i], F[i]) where S[i] is start time of meeting i and F[i] is finish time of meeting i.
What is the maximum number of meetings that can be accommodated in the meeting room?
Input:
The first line of input consists number of the test cases. The description of T test cases is as follows:
The first line consists of the size of the array, second line has the array containing the starting time of all the meetings each separated by a space, i.e., S [ i ].
And the third line has the array containing the finishing time of all the meetings each separated by a space, i.e., F [ i ].
Output:
In each separate line print the order in which the meetings take place separated by a space.
Constraints:
1 ≤ T ≤ 70
1 ≤ N ≤ 100
1 ≤ S[ i ], F[ i ] ≤ 100000
Example:
Input:
2
6
1 3 0 5 8 5
2 4 6 7 9 9
8
75250 50074 43659 8931 11273 27545 50879 77924
112960 114515 81825 93424 54316 35533 73383 160252  
Output:
1 2 4 5
6 7 1 


SOLUTION:

#include <bits/stdc++.h>
using namespace std;

int main() {
	int t;
	cin>>t;
	
	while(t--)
	{
	    int n,i,last;
	    cin>>n;
	    
	    vector<pair<int,int>> v(n);
	    unordered_map<int,int> m;
	    vector<int> s;
	    
	    for(i=0;i<n;i++)
	    cin>>v[i].second;
	    
	    for(i=0;i<n;i++)
	    {
	        cin>>v[i].first;
	        m[v[i].first+v[i].second]=i+1;
	    }     
	    
	    sort(v.begin(),v.end());
	    s.push_back(m[v[0].first+v[0].second]);
	    last=v[0].first;
	    
	    for(i=0;i<n-1;i++)
	    {
	        if(v[i+1].second>last)
	        {
	            s.push_back(m[v[i+1].first+v[i+1].second]);
	            last=v[i+1].first;
	        }
	    }
	    
	    for(auto i:s)
	    cout<<i<<" ";
	    
	    cout<<endl;
	}
	return 0;
}