LeetCode1029. Two City Scheduling

一、题目

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:

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

二、题解

class Solution {
public:
    int twoCitySchedCost(vector<vector<int>>& costs) {
        int n = costs.size();
        //每个人从a改为b的代价
        vector<int> arr(n,0);
        int sum = 0;
        for(int i = 0;i < n;i++){
            arr[i] = costs[i][1] - costs[i][0];
            sum += costs[i][0];
        }
        sort(arr.begin(),arr.end());
        for(int i = 0;i < n / 2;i++){
            sum += arr[i];
        }
        return sum;
    }
};