1029_Two_City_Scheduling.cpp

/*
A company is planning to interview 2n people. Given the array costs where costs[i] = [aCosti, bCosti], the cost of flying the ith person to city a is aCosti, and the cost of flying the ith person to city b is bCosti.

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

 

Example 1:

Input: costs = [[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.
Example 2:

Input: costs = [[259,770],[448,54],[926,667],[184,139],[840,118],[577,469]]
Output: 1859
Example 3:

Input: costs = [[515,563],[451,713],[537,709],[343,819],[855,779],[457,60],[650,359],[631,42]]
Output: 3086
 

Constraints:

2n == costs.length
2 <= costs.length <= 100
costs.length is even.
1 <= aCosti, bCosti <= 1000


来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/two-city-scheduling
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
*/

#include <vector>
#include <algorithm>
using namespace std;

bool mycmp(vector<int>& lhs, vector<int>& rhs){
	return lhs[0]-lhs[1] < rhs[0]-rhs[1];
}

class Solution {
public:
    int twoCitySchedCost(vector<vector<int>>& costs) {
		std::sort(costs.begin(), costs.end(), mycmp);
		int res = 0;
		for (size_t i = 0; i < costs.size(); ++i) {
			if(i<(costs.size()>>1)){
				res += costs[i][0];
			}
			else{
				res += costs[i][1];
			}
		}
		return res;
    }
};