[Swift]LeetCode1168. 水资源分配优化 | Optimize Water Distribution in a Village

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➤微信公众号：山青咏芝（shanqingyongzhi）
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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]

---

村里面一共有 n 栋房子。我们希望通过建造水井和铺设管道来为所有房子供水。

对于每个房子 i，我们有两种可选的供水方案：

• 一种是直接在房子内建造水井，成本为 wells[i]；
• 另一种是从另一口井铺设管道引水，数组 pipes 给出了在房子间铺设管道的成本，其中每个 pipes[i] = [house1, house2, cost] 代表用管道将 house1 和 house2 连接在一起的成本。当然，连接是双向的。

请你帮忙计算为所有房子都供水的最低总成本。

示例：

输入：n = 3, wells = [1,2,2], pipes = [[1,2,1],[2,3,1]]
输出：3
解释： 
上图展示了铺设管道连接房屋的成本。
最好的策略是在第一个房子里建造水井（成本为 1），然后将其他房子铺设管道连起来（成本为 2），所以总成本为 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

class Solution {
    func minCostToSupplyWater(_ n: Int, _ wells: [Int], _ pipes: [[Int]]) -> Int {
        var pipes = pipes
        var ds:DJSet = DJSet(n + 2)
        var m:Int = pipes.count
        pipes += [[Int]](repeating:[Int](),count:n)        
        for i in 0..<n
        {
            pipes[m+i] = [n+1, i+1, wells[i]]
        }
        pipes.sort(by:{
            $0[2] < $1[2]
        })
        var ans:Int = 0
        for e in pipes
        {
            if !ds.union(e[0], e[1])
            {
                ans += e[2]
            }
        }
        return ans
    }
}

public class DJSet
{
    var upper:[Int]
    
    init(_ n:Int)
    {
        upper = [Int](repeating:-1,count:n)
    }
    
    func root(_ x:Int) -> Int
    {
        if(upper[x] < 0)
        {
            return x
        }
        else
        {
            upper[x] = root(upper[x])
            return upper[x]
        }
    }
    
    func equiv(_ x:Int,_ y:Int) -> Bool
    {
        return root(x) == root(y)
    }
    
    func union(_ x:Int,_ y:Int) -> Bool
    {
        var x:Int = root(x)
        var y:Int = root(y)
        if x != y
        {
            if upper[y] < upper[x]
            {
                var d:Int = x
                x = y
                y = d
            }
            upper[x] += upper[y]
            upper[y] = x
        }
        return x == y
    }
    
    func count() -> Int
    {
        var ct:Int = 0
        for u in upper
        {
            if u < 0
            {
                ct += 1
            }
        }
        return ct
    }
}