[Swift]LeetCode1135. 最低成本联通所有城市 | Connecting Cities With Minimum Cost

水深无声 2021-11-17 06:12 296阅读 0赞

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★  
➤微信公众号：山青咏芝（shanqingyongzhi）  
➤博客园地址：山青咏芝（[https://www.cnblogs.com/strengthen/][https_www.cnblogs.com_strengthen]）  
➤GitHub地址：[https://github.com/strengthen/LeetCode][https_github.com_strengthen_LeetCode]  
➤原文地址：[https://www.cnblogs.com/strengthen/p/11258422.html][https_www.cnblogs.com_strengthen_p_11258422.html]   
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。  
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！  
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

There are `N` cities numbered from 1 to `N`.

You are given `connections`, where each `connections[i] = [city1, city2, cost]` represents the cost to connect `city1` and `city2` together. (A *connection* is bidirectional: connecting `city1` and `city2`is the same as connecting `city2` and `city1`.)

Return the minimum cost so that for every pair of cities, there exists a path of connections (possibly of length 1) that connects those two cities together. The cost is the sum of the connection costs used. If the task is impossible, return -1.

Example 1:

![1314_ex2.png][]

Input: N = 3, connections = [[1,2,5],[1,3,6],[2,3,1]]
    Output: 6
    Explanation: 
    Choosing any 2 edges will connect all cities so we choose the minimum 2.

Example 2:

![1314_ex1.png][]

Input: N = 4, connections = [[1,2,3],[3,4,4]]
    Output: -1
    Explanation: 
    There is no way to connect all cities even if all edges are used.

Note:

1.  `1 <= N <= 10000`
2.  `1 <= connections.length <= 10000`
3.  `1 <= connections[i][0], connections[i][1] <= N`
4.  `0 <= connections[i][2] <= 10^5`
5.  `connections[i][0] != connections[i][1]`

--------------------

想象一下你是个城市基建规划者，地图上有 `N` 座城市，它们按以 `1` 到 `N` 的次序编号。

给你一些可连接的选项 `conections`，其中每个选项 `conections[i] = [city1, city2, cost]` 表示将城市 `city1` 和城市 `city2` 连接所要的成本。（连接是双向的，也就是说城市 `city1` 和城市 `city2` 相连也同样意味着城市 `city2` 和城市 `city1` 相连）。

返回使得每对城市间都存在将它们连接在一起的连通路径（可能长度为 1 的）最小成本。该最小成本应该是所用全部连接代价的综合。如果根据已知条件无法完成该项任务，则请你返回 -1。

示例 1：

![1314_ex2.png 1][]

输入：N = 3, conections = [[1,2,5],[1,3,6],[2,3,1]]
    输出：6
    解释：
    选出任意 2 条边都可以连接所有城市，我们从中选取成本最小的 2 条。

示例 2：

![1314_ex1.png 1][]

输入：N = 4, conections = [[1,2,3],[3,4,4]]
    输出：-1
    解释： 
    即使连通所有的边，也无法连接所有城市。

提示：

1.  `1 <= N <= 10000`
2.  `1 <= conections.length <= 10000`
3.  `1 <= conections[i][0], conections[i][1] <= N`
4.  `0 <= conections[i][2] <= 10^5`
5.  `conections[i][0] != conections[i][1]`

转载于:https://www.cnblogs.com/strengthen/p/11258422.html

[https_www.cnblogs.com_strengthen]: https://www.cnblogs.com/strengthen/
[https_github.com_strengthen_LeetCode]: https://github.com/strengthen/LeetCode
[https_www.cnblogs.com_strengthen_p_11258422.html]: https://www.cnblogs.com/strengthen/p/11258422.html
[1314_ex2.png]: https://assets.leetcode.com/uploads/2019/04/20/1314_ex2.png
[1314_ex1.png]: https://assets.leetcode.com/uploads/2019/04/20/1314_ex1.png
[1314_ex2.png 1]: /images/20211116/2788a83f30a54c178385600d83766251.png
[1314_ex1.png 1]: /images/20211116/31fc3d821a6845019c03d2440ce29b06.png