1029. Two City Scheduling

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

这道题说是一个公司要面试 2n 个人,每个人飞到城市A和城市B的花费不同,现在分别让n个人去城市A和城市B面试,问最小的花费是多少。博主二话不说,上来就写个递归的暴力搜索,遍历所有的情况,但是很不幸的超时了,这道题需要更优化的解法。其实这道题可以用贪婪算法来做,首先假设我们让所有的人都去城市A,那么总花费就是把所有人去城市A的花费加起来,但现在需要让其中的一半人去城市B,由于花费不同了,怎么算呢?若去城市B的花费小于城市A的,则应该 refund 二者的差值,若去城市A的花费小于城市B的,则应该加上二者的差值。所以用去城市B的花费减去城市A的,若为负数,则是 refund,若为正数,则是追加的花费。当然是希望是负数,而且越小越好,这样就可以 refund,使得整个花费变小。所以开始时遍历一遍 costs 数组,将去城市A的花费先累加到结果 res 中,然后将去城市B的花费减去城市A的花费的差值存入 refund 数组,之后给 refund 数组排序,取出前n个值加到结果 res 中即可,参见代码如下:

class Solution {
public:
    int twoCitySchedCost(vector<vector<int>>& costs) {
        int res = 0, n = costs.size() / 2;
        vector<int> refund;
        for (auto &cost : costs) {
            res += cost[0];
            refund.push_back(cost[1] - cost[0]);
        }
        sort(refund.begin(), refund.end());
        for (int i = 0; i < n; ++i) {
            res += refund[i];
        }
        return res;
    }
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1029

参考资料:

https://leetcode.com/problems/two-city-scheduling/

https://leetcode.com/problems/two-city-scheduling/discuss/278716/C%2B%2B-O(n-log-n)-sort-by-savings

https://leetcode.com/problems/two-city-scheduling/discuss/667786/Java-or-C%2B%2B-or-Python3-or-With-detailed-explanation

LeetCode All in One 题目讲解汇总(持续更新中…)


转载请注明来源于 Grandyang 的博客 (grandyang.com),欢迎对文章中的引用来源进行考证,欢迎指出任何有错误或不够清晰的表达。可以在下面评论区评论,也可以邮件至 grandyang@qq.com

💰


微信打赏


Venmo 打赏

×

Help us with donation