[Swift]LeetCode749. 隔離病毒 | Contain Virus

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A virus is spreading rapidly, and your task is to quarantine the infected area by installing walls.

The world is modeled as a 2-D array of cells, where 0represents uninfected cells, and 1 represents cells contaminated with the virus. A wall (and only one wall) can be installed between any two 4-directionally adjacent cells, on the shared boundary.

Every night, the virus spreads to all neighboring cells in all four directions unless blocked by a wall. Resources are limited. Each day, you can install walls around only one region -- the affected area (continuous block of infected cells) that threatens the most uninfected cells the following night. There will never be a tie.

Can you save the day? If so, what is the number of walls required? If not, and the world becomes fully infected, return the number of walls used. 

Example 1:

Input: grid = 
[[0,1,0,0,0,0,0,1],
 [0,1,0,0,0,0,0,1],
 [0,0,0,0,0,0,0,1],
 [0,0,0,0,0,0,0,0]]
Output: 10
Explanation:
There are 2 contaminated regions.
On the first day, add 5 walls to quarantine the viral region on the left. The board after the virus spreads is:

[[0,1,0,0,0,0,1,1],
 [0,1,0,0,0,0,1,1],
 [0,0,0,0,0,0,1,1],
 [0,0,0,0,0,0,0,1]]

On the second day, add 5 walls to quarantine the viral region on the right. The virus is fully contained. 

Example 2:

Input: grid = 
[[1,1,1],
 [1,0,1],
 [1,1,1]]
Output: 4
Explanation: Even though there is only one cell saved, there are 4 walls built.
Notice that walls are only built on the shared boundary of two different cells. 

Example 3:

Input: grid = 
[[1,1,1,0,0,0,0,0,0],
 [1,0,1,0,1,1,1,1,1],
 [1,1,1,0,0,0,0,0,0]]
Output: 13
Explanation: The region on the left only builds two new walls. 

Note:

1. The number of rows and columns of gridwill each be in the range [1, 50].
2. Each grid[i][j] will be either 0 or 1.
3. Throughout the described process, there is always a contiguous viral region that will infect strictly more uncontaminated squares in the next round.

---

病毒擴散得很快，如今你的任務是儘量地經過安裝防火牆來隔離病毒。

假設世界由二維矩陣組成，0 表示該區域未感染病毒，而 1 表示該區域已感染病毒。能夠在任意 2 個四方向相鄰單元之間的共享邊界上安裝一個防火牆（而且只有一個防火牆）。

天天晚上，病毒會從被感染區域向相鄰未感染區域擴散，除非被防火牆隔離。現因爲資源有限，天天你只能安裝一系列防火牆來隔離其中一個被病毒感染的區域（一個區域或連續的一片區域），且該感染區域對未感染區域的威脅最大且保證惟一。

你須要努力使得最後有部分區域不被病毒感染，若是能夠成功，那麼返回須要使用的防火牆個數; 若是沒法實現，則返回在世界被病毒所有感染時已安裝的防火牆個數。 

示例 1：

輸入: grid = 
[[0,1,0,0,0,0,0,1],
 [0,1,0,0,0,0,0,1],
 [0,0,0,0,0,0,0,1],
 [0,0,0,0,0,0,0,0]]
輸出: 10
說明:
一共有兩塊被病毒感染的區域: 從左往右第一塊須要 5 個防火牆，同時若該區域不隔離，晚上將感染 5 個未感染區域（即被威脅的未感染區域個數爲 5）;
第二塊須要 4 個防火牆，同理被威脅的未感染區域個數是 4。所以，第一天先隔離左邊的感染區域，通過一晚後，病毒傳播後世界以下:
[[0,1,0,0,0,0,1,1],
 [0,1,0,0,0,0,1,1],
 [0,0,0,0,0,0,1,1],
 [0,0,0,0,0,0,0,1]]
第二題，只剩下一塊未隔離的被感染的連續區域，此時須要安裝 5 個防火牆，且安裝完畢後病毒隔離任務完成。

示例 2：

輸入: grid = 
[[1,1,1],
 [1,0,1],
 [1,1,1]]
輸出: 4
說明: 
此時只須要安裝 4 面防火牆，就有一小區域能夠倖存，不被病毒感染。
注意不須要在世界邊界創建防火牆。 

示例 3:

輸入: grid = 
[[1,1,1,0,0,0,0,0,0],
 [1,0,1,0,1,1,1,1,1],
 [1,1,1,0,0,0,0,0,0]]
輸出: 13
說明: 
在隔離右邊感染區域後，隔離左邊病毒區域只須要 2 個防火牆了。 

說明:

1. grid 的行數和列數範圍是 [1, 50]。
2.  grid[i][j] 只包含 0 或 1 。
3. 題目保證每次選取感染區域進行隔離時，必定存在惟一一個對未感染區域的威脅最大的區域。

---

Runtime: 92 ms

Memory Usage: 19.5 MB

 1 class Solution {
 2     func containVirus(_ grid: [[Int]]) -> Int {
 3         var grid = grid
 4         var res:Int = 0
 5         var m:Int = grid.count
 6         var n:Int = grid[0].count
 7         var dirs:[[Int]] = [[-1,0],[0,1],[1,0],[0,-1]]
 8         while (true )
 9         {
10             var visited:Set<Int> = Set<Int>()
11             var all:[[[Int]]] = [[[Int]]]()
12             for i in 0..<m
13             {
14                 for j in 0..<n
15                 {
16                     if grid[i][j] == 1 && !visited.contains(i * n + j)
17                     {
18                         var q:[Int] = [i * n + j]
19                         var virus:[Int] = [i * n + j]
20                         var walls:[Int] = [Int]()
21                         visited.insert(i * n + j)
22                         while(!q.isEmpty)
23                         {
24                             var t:Int = q.removeFirst()                            
25                             for dir in dirs
26                             {
27                                 var x:Int = (t / n) + dir[0]
28                                 var y:Int = (t % n) + dir[1]
29                                 if x < 0 || x >= m || y < 0 || y >= n || visited.contains(x * n + y)
30                                 {
31                                     continue
32                                 }
33                                 if grid[x][y] == -1
34                                 {
35                                     continue
36                                 }
37                                 else if grid[x][y] == 0
38                                 {
39                                     walls.append(x * n + y)
40                                 }
41                                 else if grid[x][y] == 1
42                                 {
43                                     visited.insert(x * n + y)
44                                     virus.append(x * n + y)
45                                     q.append(x * n + y)
46                                 }
47                             }                            
48                         }   
49                         var s:Set<Int> = Set(walls)
50                         var cells:[Int] = [s.count]
51                         all.append([cells ,walls, virus])
52                     }
53                 }
54             }
55             if all.isEmpty {break}
56             all.sort(by:{(a:[[Int]],b:[[Int]]) -> Bool in 
57                         return a[0][0] > b[0][0]})
58             for i in 0..<all.count
59             {
60                 if i == 0
61                 {
62                     var virus:[Int] = all[0][2]
63                     for idx in virus
64                     {
65                         grid[idx / n][idx % n] = -1
66                     }
67                     res += all[0][1].count
68                 }
69                 else
70                 {
71                     var wall:[Int] = all[i][1]
72                     for idx in wall
73                     {
74                         grid[idx / n][idx % n] = 1
75                     }
76                 }
77             }     
78         }
79         return res
80     }
81 }