3.Two-City scheduling.cpp

/*
There are 2N people a company is planning to interview. The cost of flying the i-th person to city A is costs[i][0], and the cost of flying the i-th person to city B is costs[i][1].

Return the minimum cost to fly every person to a city such that exactly N people arrive in each city.



Example 1:

Input: [[10,20],[30,200],[400,50],[30,20]]
Output: 110
Explanation:
The first person goes to city A for a cost of 10.
The second person goes to city A for a cost of 30.
The third person goes to city B for a cost of 50.
The fourth person goes to city B for a cost of 20.

The total minimum cost is 10 + 30 + 50 + 20 = 110 to have half the people interviewing in each city.
*/

class Solution {
public:
    int twoCitySchedCost(vector<vector<int>>& costs) {
        int n=costs.size();
        priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>> pq;
        for(int i=0;i<n;i++){
            int d=costs[i][0]-costs[i][1];
            pq.push(make_pair(d,costs[i][0]));
        }
        int cost=0;
        for(int i=1;i<=n/2;i++){
            int c=pq.top().second;
            cost+=c;
            pq.pop();
        }
           for(int i=1;i<=n/2;i++){
            int c=pq.top().first;
            cost+=(pq.top().second-c);
            pq.pop();
        }
        return cost;
    }
};