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787-cheapest-flights-within-k-stops.md Added Python solution.
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#787. Cheapest Flights Within K Stops - Best Practices of LeetCode Solutions
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LeetCode link:[787. Cheapest Flights Within K Stops](https://leetcode.com/problems/cheapest-flights-within-k-stops), difficulty:**Medium**.
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##LeetCode description of "787. Cheapest Flights Within K Stops"
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There are`n` cities connected by some number of flights. You are given an array`flights` where`flights[i] = [from_i, to_i, price_i]` indicates that there is a flight from city`from_i` to city`to_i` with cost`price_i`.
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You are also given three integers`src`,`dst`, and`k`, return_**the cheapest price** from`src` to`dst` with at most`k` stops_. If there is no such route, return`-1`.
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###[Example 1]
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![](../../images/examples/787_1.png)
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**Input**:`n = 4, flights = [[0,1,100],[1,2,100],[2,0,100],[1,3,600],[2,3,200]], src = 0, dst = 3, k = 1`
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**Output**:`700`
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**Explanation**:
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```
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The graph is shown above.
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The optimal path with at most 1 stop from city 0 to 3 is marked in red and has cost 100 + 600 = 700.
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Note that the path through cities [0,1,2,3] is cheaper but is invalid because it uses 2 stops.
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```
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###[Example 2]
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**Input**:`n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 1`
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**Output**:`200`
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**Explanation**:
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```
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The graph is shown above.
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The optimal path with at most 1 stop from city 0 to 2 is marked in red and has cost 100 + 100 = 200.
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```
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###[Example 3]
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**Input**:`n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 0`
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**Output**:`500`
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**Explanation**:
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```
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The graph is shown above.
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The optimal path with no stops from city 0 to 2 is marked in red and has cost 500.
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```
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###[Constraints]
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-`1 <= n <= 100`
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-`0 <= flights.length <= (n * (n - 1) / 2)`
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-`flights[i].length == 3`
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-`0 <= from_i, to_i < n`
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-`from_i != to_i`
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-`1 <= price_i <= 10000`
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- There will not be any multiple flights between two cities.
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-`0 <= src, dst, k < n`
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-`src != dst`
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##Intuition
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We can solve it via**Bellman-Ford algorithm**.
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For a detailed introduction to`Bellman-Ford algorithm`, please refer to[743. Network Delay Time](./743-network-delay-time.md).
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##Complexity
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**V**: vertex count,**E**: Edge count.
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* Time:`O(K * E)`.
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* Space:`O(V)`.
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##Python
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```python
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classSolution:
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deffindCheapestPrice(self,n:int,flights: List[List[int]],src:int,dst:int,k:int) ->int:
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min_costs= [float('inf')]* n
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min_costs[src]=0
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for _inrange(k+1):
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min_costs_clone= min_costs.copy()
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for from_city, to_city, pricein flights:
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if min_costs_clone[from_city]==float('inf'):
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continue
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cost= min_costs_clone[from_city]+ price
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if cost< min_costs[to_city]:
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min_costs[to_city]= cost
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if min_costs[dst]==float('inf'):
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return-1
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return min_costs[dst]
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```
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##JavaScript
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```javascript
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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##Java
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```java
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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##C++
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```cpp
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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##C#
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```c#
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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##Go
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```go
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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##Ruby
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```ruby
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# Welcome to create a PR to complete the code of this language, thanks!
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```
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##C, Kotlin, Swift, Rust or other languages
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```
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// Welcome to create a PR to complete the code of this language, thanks!
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```

‎images/examples/787_1.png

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‎images/examples/787_2.png

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‎images/examples/787_3.png

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#787. K 站中转内最便宜的航班 - 力扣题解最佳实践
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力扣链接：[787. K 站中转内最便宜的航班](https://leetcode.cn/problems/cheapest-flights-within-k-stops)，难度:**中等**。
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##力扣“787. K 站中转内最便宜的航班”问题描述
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有`n` 个城市通过一些航班连接。给你一个数组`flights` ，其中`flights[i] = [fromi, toi, pricei]` ，表示该航班都从城市`fromi` 开始，以价格`pricei` 抵达`toi`。
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现在给定所有的城市和航班，以及出发城市`src` 和目的地`dst`，你的任务是找到出一条最多经过`k` 站中转的路线，使得从`src` 到`dst` 的**价格最便宜** ，并返回该价格。 如果不存在这样的路线，则输出`-1`。
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###[示例 1]
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![](../../images/examples/787_1.png)
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**输入**:`n = 4, flights = [[0,1,100],[1,2,100],[2,0,100],[1,3,600],[2,3,200]], src = 0, dst = 3, k = 1`
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**输出**:`700`
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**解释**:
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```
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城市航班图如上
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从城市 0 到城市 3 经过最多 1 站的最佳路径用红色标记，费用为 100 + 600 = 700。
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请注意，通过城市 [0, 1, 2, 3] 的路径更便宜，但无效，因为它经过了 2 站。
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```
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###[示例 2]
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**输入**:`n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 1`
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**输出**:`200`
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**解释**:
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```
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城市航班图如上
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从城市 0 到城市 2 经过最多 1 站的最佳路径标记为红色，费用为 100 + 100 = 200。
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```
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###[示例 3]
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**输入**:`n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 0`
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**输出**:`500`
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**解释**:
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```
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城市航班图如上
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从城市 0 到城市 2 不经过站点的最佳路径标记为红色，费用为 500。
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```
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###[约束]
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-`1 <= n <= 100`
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-`0 <= flights.length <= (n * (n - 1) / 2)`
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-`flights[i].length == 3`
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-`0 <= from_i, to_i < n`
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-`from_i != to_i`
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-`1 <= price_i <= 10000`
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- 航班没有重复，且不存在自环
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-`0 <= src, dst, k < n`
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-`src != dst`
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##思路
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本题可用**Bellman-Ford算法** 解决。
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**Bellman-Ford算法**的详细说明，请参考[743. 网络延迟时间](./743-network-delay-time.md)。
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##复杂度
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**V**: 顶点数量，**E**: 边的数量。
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* 时间:`O(K * E)`.
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* 空间:`O(V)`.
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##Python
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```python
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classSolution:
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deffindCheapestPrice(self,n:int,flights: List[List[int]],src:int,dst:int,k:int) ->int:
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min_costs= [float('inf')]* n
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min_costs[src]=0
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for _inrange(k+1):
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min_costs_clone= min_costs.copy()
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for from_city, to_city, pricein flights:
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if min_costs_clone[from_city]==float('inf'):
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continue
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cost= min_costs_clone[from_city]+ price
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if cost< min_costs[to_city]:
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min_costs[to_city]= cost
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if min_costs[dst]==float('inf'):
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return-1
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return min_costs[dst]
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```
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##JavaScript
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```javascript
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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##Java
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```java
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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##C++
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```cpp
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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##C#
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```c#
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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##Go
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```go
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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##Ruby
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```ruby
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# Welcome to create a PR to complete the code of this language, thanks!
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```
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##C, Kotlin, Swift, Rust or other languages
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```
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// Welcome to create a PR to complete the code of this language, thanks!
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```

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