原文链接: https://leetcode-cn.com/problems/two-city-scheduling

英文原文

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

中文题目

公司计划面试 2n 人。给你一个数组 costs ，其中 costs[i] = [aCosti, bCosti] 。第 i 人飞往 a 市的费用为 aCosti ，飞往 b 市的费用为 bCosti 。

返回将每个人都飞到 a 、b 中某座城市的最低费用，要求每个城市都有 n 人抵达。

 

示例 1：

输入：costs = [[10,20],[30,200],[400,50],[30,20]]
输出：110
解释：
第一个人去 a 市，费用为 10。
第二个人去 a 市，费用为 30。
第三个人去 b 市，费用为 50。
第四个人去 b 市，费用为 20。

最低总费用为 10 + 30 + 50 + 20 = 110，每个城市都有一半的人在面试。

示例 2：

输入：costs = [[259,770],[448,54],[926,667],[184,139],[840,118],[577,469]]
输出：1859

示例 3：

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

 

提示：

• 2 * n == costs.length
• 2 <= costs.length <= 100
• costs.length 为偶数
• 1 <= aCosti, bCosti <= 1000

通过代码

官方题解

方法一：贪心

分析

我们这样来看这个问题，公司首先将这 $2N$ 个人全都安排飞往 $B$ 市，再选出 $N$ 个人改变它们的行程，让他们飞往 $A$ 市。如果选择改变一个人的行程，那么公司将会额外付出 price_A - price_B 的费用，这个费用可正可负。

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因此最优的方案是，选出 price_A - price_B 最小的 $N$ 个人，让他们飞往 A 市，其余人飞往 B 市。

算法

• 按照 price_A - price_B 从小到大排序；

• 将前 $N$ 个人飞往 A 市，其余人飞往 B 市，并计算出总费用。

[sol1]
class Solution:
    def twoCitySchedCost(self, costs: List[List[int]]) -> int:
        # Sort by a gain which company has 
        # by sending a person to city A and not to city B
        costs.sort(key = lambda x : x[0] - x[1])
        
        total = 0
        n = len(costs) // 2
        # To optimize the company expenses,
        # send the first n persons to the city A
        # and the others to the city B
        for i in range(n):
            total += costs[i][0] + costs[i + n][1]
        return total

[sol1]
class Solution {
    public int twoCitySchedCost(int[][] costs) {
      // Sort by a gain which company has 
      // by sending a person to city A and not to city B
      Arrays.sort(costs, new Comparator<int[]>() {
          @Override
          public int compare(int[] o1, int[] o2) {
              return o1[0] - o1[1] - (o2[0] - o2[1]);
          }
      });

      int total = 0;
      int n = costs.length / 2;
      // To optimize the company expenses,
      // send the first n persons to the city A
      // and the others to the city B
      for (int i = 0; i < n; ++i) total += costs[i][0] + costs[i + n][1];
      return total;
    }
}

[sol1]
class Solution {
    public:
    int twoCitySchedCost(vector<vector<int>>& costs) {
        // Sort by a gain which company has 
        // by sending a person to city A and not to city B
        sort(begin(costs), end(costs),
                [](const vector<int> &o1, const vector<int> &o2) {
            return (o1[0] - o1[1] < o2[0] - o2[1]);
        });

        int total = 0;
        int n = costs.size() / 2;
        // To optimize the company expenses,
        // send the first n persons to the city A
        // and the others to the city B
        for (int i = 0; i < n; ++i) total += costs[i][0] + costs[i + n][1];
        return total;
    }
};

复杂度分析

• 时间复杂度：$O(N \log N)$，需要对 price_A - price_B 进行排序。

• 空间复杂度：$O(1)$。

统计信息

通过次数	提交次数	AC比率
17489	26532	65.9%

提交历史

提交时间	提交结果	执行时间	内存消耗	语言