HackerEarth Pro and Con List problem solution

In this HackerEarth Pro and Con List problem solution, There is a long list of n girls in front of Barney, and he is to calculate the optimal “happiness” he can find by selecting exactly 2 girls. (Why 2? No one knows!)

Ted, as a fan of pros and cons, suggests to make a list, a method for estimating the maximum happiness that Barney can achieve.

Each girl is characterized by two parameters:

– favour: if this girl is chosen, his happiness increases by this amount.

– anger: if this girl is not chosen, his happiness decreases by this amount.

Find the maximum “happiness” that Barney can obtain. Note that the answer is allowed to be negative.

#include <bits/stdc++.h>

using namespace std;

int main ()
{
    int tc;
    
    vector < int > favor;
    vector < int > anger;
    vector < long long int > ans;
    
    scanf("%d",&tc);

    assert(1<=tc);
    assert(tc<=10);

    while (tc--)
    {
        int n;

        favor.clear();
        anger.clear();
        ans.clear();
        
        scanf("%d",&n);
        assert(2<=n);
        assert(n<=100000);
        
        favor.resize(n);
        anger.resize(n);
        ans.resize(n);
        
        long long int hatao = 0;

        for (int i=0; i<n; i++)
        {
            scanf("%d %d",&favor[i], &anger[i]);

            ans[i] = favor[i] + anger[i];
            hatao = hatao + anger[i];
            
            assert(0<=favor[i]);
            assert(favor[i]<=1000000000);
            assert(0<=anger[i]);
            assert(anger[i]<=1000000000);
        }

        sort(ans.rbegin(),ans.rend());

        cout << ans[0] + ans[1] - hatao << endl;
    }
    return 0;
}