In this HackerEarth [Only for Data Science] - Bank Happiness Index problem solution you are given data of customers that visit a bank in a day. The data contains the following details: (Customer ID, Entry time, Maximum wait time, Happiness Factor, time required for work completion). There N are such customers. There are only K members in the staff that can assist these customers. Given the data, you need to assign these K staff members to the customers in such a way that the total happiness at the end of the day is as maximum possible. There are the following constraints of assigning the staff members - 

  1. You can assign a staff member to at most one person at a time.
  2. The closing time of the bank is 5:30 PM i.e. (1050th minute of the day). The time allocation should not be greater than that. Also, you need to allocate time in such a way that its completion time from the start does not exceed 1050.
  3. The wait time of a particular customer includes its work completion time. Also, the customer remains in the bank even if his wait time gets over till 1050th minute of the day.


hackerEarth [Only for Data Science] - Bank Happiness Index problem solution


hackerEarth [Only for Data Science] - Bank Happiness Index problem solution.

#include<bits/stdc++.h>
#define LL long long int
#define M 1000000007
#define endl "\n"
#define eps 0.00000001
LL pow(LL a,LL b,LL m){ a%=m;LL x=1,y=a;while(b > 0){if(b%2 == 1){x=(x*y);if(x>m) x%=m;}y = (y*y);if(y>m) y%=m;b /= 2;}return x%m;}
LL gcd(LL a,LL b){if(b==0) return a; else return gcd(b,a%b);}
LL gen(LL start,LL end){LL diff = end-start;LL temp = rand()%start;return temp+diff;}
using namespace std;
LL id[100001], entry[100001], wait[100001], req_time[100001];
double happiness[100001];
LL allocation_time[100001], staff_id[100001];
LL busy[100001];
int main()  {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n, k;
    assert(cin >> n >> k);
    assert(n >= 1 && n <= 100000);
    assert(k >= 1 && k <= 100000);
    for(int i = 1; i <= n; i++) {
        cin >> id[i] >> entry[i] >> wait[i] >> happiness[i] >> req_time[i];
        assert(id[i] >= 0 && id[i] <= 1e9);
        assert(wait[i] >= req_time[i] && wait[i] <= 180);
    }
    double score = 0.0;
    vector<pair<pair<int,int> , int> > status;
    for(int i = 1; i <= n; i++) {
        cin >> allocation_time[i] >> staff_id[i];
        if(allocation_time[i] == -1 || staff_id[i] == -1) {
            assert(allocation_time[i] == staff_id[i]);
            score = score + (entry[i] - 1050) * happiness[i] * 0.1;
            continue;
        }
        assert(staff_id[i] >= 1 && staff_id[i] <= k);
        assert(allocation_time[i] >= entry[i] && allocation_time[i] <= 1050 - req_time[i]);
        score = score + (entry[i] + wait[i] - allocation_time[i] - req_time[i]) * happiness[i] * 100;
        status.push_back({{allocation_time[i] , staff_id[i]} , i });
    }
    sort(status.begin() , status.end());
    for(int i = 0; i < status.size(); i++) {
        int staff_id = status[i].first.second;
        int alloc_time = status[i].first.first;
        int user_id = status[i].second;
        assert(busy[staff_id] <= alloc_time);
        busy[staff_id] = alloc_time + req_time[user_id]; 
    }
    cout << (score+(long long)(5e6));
}