[Swift]LeetCode1066. 校園自行車分配 II | Campus Bikes II

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➤微信公眾號：山青詠芝（shanqingyongzhi）
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On a campus represented as a 2D grid, there are N workers and M bikes, with N <= M. Each worker and bike is a 2D coordinate on this grid.

We assign one unique bike to each worker so that the sum of the Manhattan distances between each worker and their assigned bike is minimized.

The Manhattan distance between two points p1 and p2 is Manhattan(p1, p2) = |p1.x - p2.x| + |p1.y - p2.y|.

Return the minimum possible sum of Manhattan distances between each worker and their assigned bike.

Example 1:

Input: workers = [[0,0],[2,1]], bikes = [[1,2],[3,3]] Output: 6 Explanation: We assign bike 0 to worker 0, bike 1 to worker 1. The Manhattan distance of both assignments is 3, so the output is 6. 

Example 2:

Input: workers = [[0,0],[1,1],[2,0]], bikes = [[1,0],[2,2],[2,1]] Output: 4 Explanation: We first assign bike 0 to worker 0, then assign bike 1 to worker 1 or worker 2, bike 2 to worker 2 or worker 1. Both assignments lead to sum of the Manhattan distances as 4.

Note:

1. 0 <= workers[i][0], workers[i][1], bikes[i][0], bikes[i][1] < 1000
2. All worker and bike locations are distinct.
3. 1 <= workers.length <= bikes.length <= 10

---

在由 2D 網格表示的校園里有 n 位工人（worker）和 m 輛自行車（bike），n <= m。所有工人和自行車的位置都用網格上的 2D 坐標表示。

我們為每一位工人分配一輛專屬自行車，使每個工人與其分配到的自行車之間的曼哈頓距離最小化。

p1 和 p2 之間的曼哈頓距離為 Manhattan(p1, p2) = |p1.x - p2.x| + |p1.y - p2.y|。

返回每個工人與分配到的自行車之間的曼哈頓距離的最小可能總和。

示例 1：

輸入：workers = [[0,0],[2,1]], bikes = [[1,2],[3,3]]
輸出：6
解釋：
自行車 0 分配給工人 0，自行車 1 分配給工人 1 。分配得到的曼哈頓距離都是 3, 所以輸出為 6 。

示例 2：

輸入：workers = [[0,0],[1,1],[2,0]], bikes = [[1,0],[2,2],[2,1]]
輸出：4
解釋：
先將自行車 0 分配給工人 0，再將自行車 1 分配給工人 1（或工人 2），自行車 2 給工人 2（或工人 1）。如此分配使得曼哈頓距離的總和為 4。

提示：

1. 0 <= workers[i][0], workers[i][1], bikes[i][0], bikes[i][1] < 1000
2. 所有工人和自行車的位置都不相同。
3. 1 <= workers.length <= bikes.length <= 10

---

Runtime: 36 ms

Memory Usage: 20.9 MB

class Solution {
    var dp:[[Int]] = [[Int]](repeating:[Int](repeating:-1,count:11),count:1 << 11)
    func assignBikes(_ workers: [[Int]], _ bikes: [[Int]]) -> Int {
        var workers = workers
        var bikes = bikes
        return F(&workers, &bikes, 0, 0)
    }
    
    func getCost(_ a:[Int],_ b:[Int]) -> Int
    {
        var ret:Int = 0
        for i in 0..<a.count
        {
            ret += abs(a[i] - b[i])
        }
        return ret
    }
    
    func F(_ w:inout [[Int]], _ b:inout [[Int]],_ bikesTaken:Int,_ workerId:Int) -> Int
    {
        var x:Int = b.count
        if workerId == w.count {return 0}
        else if bikesTaken + 1 == (1 << x) {return 0}
        else if dp[bikesTaken][workerId] != -1 {return dp[bikesTaken][workerId]}
        
        var curr:Int = 1000000000
        for i in 0..<x
        {
            if (bikesTaken & (1 << i)) != 0 {continue}
            curr = min(curr, F(&w, &b, bikesTaken | (1 << i), workerId + 1) + getCost(b[i], w[workerId]))
        }
        dp[bikesTaken][workerId] = curr
        return curr
    }
}

---

回溯法：Time Limit Exceeded

class Solution {
    var ans:Int = -1
    var vis:[Bool] = [Bool](repeating:false,count:20)
    func assignBikes(_ workers: [[Int]], _ bikes: [[Int]]) -> Int {
        var list1:[Position] = [Position]()
        var list2:[Position] = [Position]()
        for pos in workers
        {
            list1.append(Position(pos[0] , pos[1]))
        }
        for pos in bikes
        {
            list2.append(Position(pos[0] , pos[1]))
        }
        backtracking(list1 , 0 , list2 , 0)
        return ans
    }
    
    func getDist(_ pos1:Position ,_ pos2:Position) -> Int
    {
        return abs(pos1.x - pos2.x) + abs(pos1.y - pos2.y)
    }
    
    func backtracking(_ list1:[Position],_ current:Int,_ list2:[Position],_ dist:Int)
    {
        if current == list1.count
        {
            if dist < ans || ans < 0 {ans = dist}
        }
        else
        {
            for i in 0..<list2.count
            {
                if !vis[i]
                {
                    vis[i] = true
                    backtracking(list1 , current + 1 , list2 , dist + getDist(list1[current] , list2[i]))
                    vis[i] = false
                }
            }
        }
    }
}

class Position
{
    var x:Int
    var y:Int
    init(_ x:Int,_ y:Int)
    {
        self.x = x
        self.y = y
    }
}

---

DFS：Time Limit Exceeded 

class Solution {
    var minNum:Int = Int.max
    func assignBikes(_ workers: [[Int]], _ bikes: [[Int]]) -> Int {
        var arrBool:[Bool] = [Bool](repeating:false,count:bikes.count)   
        dfs(workers, 0, bikes,&arrBool, 0)
        return minNum
    }
    
    func dfs(_ workers: [[Int]],_ i:Int,_ bikes: [[Int]],_ used:inout [Bool],_ sum:Int)
    {
        if i == workers.count
        {
            minNum = min(minNum, sum);
            return
        }
        
        for j in 0..<bikes.count
        {
            if used[j] {continue}
            used[j] = true
            dfs(workers, i+1, bikes, &used, sum + getDistance(workers[i], bikes[j]))
            used[j] = false
        }
    }
    
    func getDistance(_ p1:[Int],_ p2:[Int]) -> Int
    {
        return abs(p1[0] - p2[0]) + abs(p1[1] - p2[1])
    }
}