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

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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 網格表示的校園裏有 n 位工人（worker）和 m 輛自行車（bike），n <= m。全部工人和自行車的位置都用網格上的 2D 座標表示。

咱們爲每一位工人分配一輛專屬自行車，使每一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 之間的曼哈頓距離爲 Manhattan(p1, p2) = |p1.x - p2.x| + |p1.y - p2.y|。

返回每一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 。分配獲得的曼哈頓距離都是 3, 因此輸出爲 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爲 4。

提示：

1. 0 <= workers[i][0], workers[i][1], bikes[i][0], bikes[i][1] < 1000
2. 全部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

 1 class Solution {
 2     var dp:[[Int]] = [[Int]](repeating:[Int](repeating:-1,count:11),count:1 << 11)
 3     func assignBikes(_ workers: [[Int]], _ bikes: [[Int]]) -> Int {
 4         var workers = workers
 5         var bikes = bikes
 6         return F(&workers, &bikes, 0, 0)
 7     }
 8     
 9     func getCost(_ a:[Int],_ b:[Int]) -> Int
10     {
11         var ret:Int = 0
12         for i in 0..<a.count
13         {
14             ret += abs(a[i] - b[i])
15         }
16         return ret
17     }
18     
19     func F(_ w:inout [[Int]], _ b:inout [[Int]],_ bikesTaken:Int,_ workerId:Int) -> Int
20     {
21         var x:Int = b.count
22         if workerId == w.count {return 0}
23         else if bikesTaken + 1 == (1 << x) {return 0}
24         else if dp[bikesTaken][workerId] != -1 {return dp[bikesTaken][workerId]}
25         
26         var curr:Int = 1000000000
27         for i in 0..<x
28         {
29             if (bikesTaken & (1 << i)) != 0 {continue}
30             curr = min(curr, F(&w, &b, bikesTaken | (1 << i), workerId + 1) + getCost(b[i], w[workerId]))
31         }
32         dp[bikesTaken][workerId] = curr
33         return curr
34     }
35 }

---

回溯法：Time Limit Exceeded

 1 class Solution {
 2     var ans:Int = -1
 3     var vis:[Bool] = [Bool](repeating:false,count:20)
 4     func assignBikes(_ workers: [[Int]], _ bikes: [[Int]]) -> Int {
 5         var list1:[Position] = [Position]()
 6         var list2:[Position] = [Position]()
 7         for pos in workers
 8         {
 9             list1.append(Position(pos[0] , pos[1]))
10         }
11         for pos in bikes
12         {
13             list2.append(Position(pos[0] , pos[1]))
14         }
15         backtracking(list1 , 0 , list2 , 0)
16         return ans
17     }
18     
19     func getDist(_ pos1:Position ,_ pos2:Position) -> Int
20     {
21         return abs(pos1.x - pos2.x) + abs(pos1.y - pos2.y)
22     }
23     
24     func backtracking(_ list1:[Position],_ current:Int,_ list2:[Position],_ dist:Int)
25     {
26         if current == list1.count
27         {
28             if dist < ans || ans < 0 {ans = dist}
29         }
30         else
31         {
32             for i in 0..<list2.count
33             {
34                 if !vis[i]
35                 {
36                     vis[i] = true
37                     backtracking(list1 , current + 1 , list2 , dist + getDist(list1[current] , list2[i]))
38                     vis[i] = false
39                 }
40             }
41         }
42     }
43 }
44 
45 class Position
46 {
47     var x:Int
48     var y:Int
49     init(_ x:Int,_ y:Int)
50     {
51         self.x = x
52         self.y = y
53     }
54 }

---

DFS：Time Limit Exceeded 

 1 class Solution {
 2     var minNum:Int = Int.max
 3     func assignBikes(_ workers: [[Int]], _ bikes: [[Int]]) -> Int {
 4         var arrBool:[Bool] = [Bool](repeating:false,count:bikes.count)   
 5         dfs(workers, 0, bikes,&arrBool, 0)
 6         return minNum
 7     }
 8     
 9     func dfs(_ workers: [[Int]],_ i:Int,_ bikes: [[Int]],_ used:inout [Bool],_ sum:Int)
10     {
11         if i == workers.count
12         {
13             minNum = min(minNum, sum);
14             return
15         }
16         
17         for j in 0..<bikes.count
18         {
19             if used[j] {continue}
20             used[j] = true
21             dfs(workers, i+1, bikes, &used, sum + getDistance(workers[i], bikes[j]))
22             used[j] = false
23         }
24     }
25     
26     func getDistance(_ p1:[Int],_ p2:[Int]) -> Int
27     {
28         return abs(p1[0] - p2[0]) + abs(p1[1] - p2[1])
29     }
30 }