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
这题一开始觉得应该用 DP 来做, 结果超时了。然后自己想了另一种解法, 感觉应该算是贪婪算法, 不知道是不是, 记录在这,大家看一下:
- 先把
costs
均分成两个数组to_a
(到城市 A 进行面试)和to_b
(到城市 B 进行面试) - 遍历
to_a
将每一个元素与to_b
中的每个元素进行对比, 如果交换之后可以减少成本则进行交换
每一步的最优解积累成最终的整体最优
impl Solution {
pub fn two_city_sched_cost(costs: Vec<Vec<i32>>) -> i32 {
let mut to_a = costs[..costs.len() / 2].to_vec();
let mut to_b = costs[costs.len() / 2..].to_vec();
for i in 0..costs.len() / 2 {
for j in 0..costs.len() / 2 {
let a = to_a[i].clone();
let b = to_b[j].clone();
if a[1] + b[0] < a[0] + b[1] {
to_a[i] = b;
to_b[j] = a;
}
}
}
to_a.into_iter().map(|v| v[0]).sum::<i32>() + to_b.into_iter().map(|v| v[1]).sum::<i32>()
}
}