[Swift]LeetCode1168. 水資源分配優化 | Optimize Water Distribution in a Village

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There are n houses in a village. We want to supply water for all the houses by building wells and laying pipes.

For each house i, we can either build a well inside it directly with cost wells[i], or pipe in water from another well to it. The costs to lay pipes between houses are given by the array pipes, where each pipes[i] = [house1, house2, cost] represents the cost to connect house1 and house2 together using a pipe. Connections are bidirectional.

Find the minimum total cost to supply water to all houses.

Example 1:

Input: n = 3, wells = [1,2,2], pipes = [[1,2,1],[2,3,1]]
Output: 3
Explanation: 
The image shows the costs of connecting houses using pipes.
The best strategy is to build a well in the first house with cost 1 and connect the other houses to it with cost 2 so the total cost is 3.

Constraints:

• 1 <= n <= 10000
• wells.length == n
• 0 <= wells[i] <= 10^5
• 1 <= pipes.length <= 10000
• 1 <= pipes[i][0], pipes[i][1] <= n
• 0 <= pipes[i][2] <= 10^5
• pipes[i][0] != pipes[i][1]

---

村裏面一共有 n 棟房子。咱們但願經過建造水井和鋪設管道來爲全部房子供水。

對於每一個房子 i，咱們有兩種可選的供水方案：

• 一種是直接在房子內建造水井，成本爲 wells[i]；
• 另外一種是從另外一口井鋪設管道引水，數組 pipes 給出了在房子間鋪設管道的成本，其中每一個 pipes[i] = [house1, house2, cost] 表明用管道將 house1 和 house2 鏈接在一塊兒的成本。固然，鏈接是雙向的。

請你幫忙計算爲全部房子都供水的最低總成本。

 

示例：

輸入：n = 3, wells = [1,2,2], pipes = [[1,2,1],[2,3,1]]
輸出：3
解釋： 
上圖展現了鋪設管道鏈接房屋的成本。
最好的策略是在第一個房子裏建造水井（成本爲 1），而後將其餘房子鋪設管道連起來（成本爲 2），因此總成本爲 3。

 

提示：

• 1 <= n <= 10000
• wells.length == n
• 0 <= wells[i] <= 10^5
• 1 <= pipes.length <= 10000
• 1 <= pipes[i][0], pipes[i][1] <= n
• 0 <= pipes[i][2] <= 10^5
• pipes[i][0] != pipes[i][1]

---

1260 ms

 1 class Solution {
 2     func minCostToSupplyWater(_ n: Int, _ wells: [Int], _ pipes: [[Int]]) -> Int {
 3         var pipes = pipes
 4         var ds:DJSet = DJSet(n + 2)
 5         var m:Int = pipes.count
 6         pipes += [[Int]](repeating:[Int](),count:n)        
 7         for i in 0..<n
 8         {
 9             pipes[m+i] = [n+1, i+1, wells[i]]
10         }
11         pipes.sort(by:{
12             $0[2] < $1[2]
13         })
14         var ans:Int = 0
15         for e in pipes
16         {
17             if !ds.union(e[0], e[1])
18             {
19                 ans += e[2]
20             }
21         }
22         return ans
23     }
24 }
25 
26 public class DJSet
27 {
28     var upper:[Int]
29     
30     init(_ n:Int)
31     {
32         upper = [Int](repeating:-1,count:n)
33     }
34     
35     func root(_ x:Int) -> Int
36     {
37         if(upper[x] < 0)
38         {
39             return x
40         }
41         else
42         {
43             upper[x] = root(upper[x])
44             return upper[x]
45         }
46     }
47     
48     func equiv(_ x:Int,_ y:Int) -> Bool
49     {
50         return root(x) == root(y)
51     }
52     
53     func union(_ x:Int,_ y:Int) -> Bool
54     {
55         var x:Int = root(x)
56         var y:Int = root(y)
57         if x != y
58         {
59             if upper[y] < upper[x]
60             {
61                 var d:Int = x
62                 x = y
63                 y = d
64             }
65             upper[x] += upper[y]
66             upper[y] = x
67         }
68         return x == y
69     }
70     
71     func count() -> Int
72     {
73         var ct:Int = 0
74         for u in upper
75         {
76             if u < 0
77             {
78                 ct += 1
79             }
80         }
81         return ct
82     }
83 }