leetcode-1029 Two City Scheduling

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>>& cs, int res = 0) {
      sort(begin(cs), end(cs), [](vector<int> &v1, vector<int> &v2) {
        return (v1[0] - v1[1] < v2[0] - v2[1]);
      });
      for (auto i = 0; i < cs.size() / 2; ++i) {
        res += cs[i][0] + cs[i + cs.size() / 2][1];
      }
      return res;
    }
};