[Swift]LeetCode749. 隔离病毒 | 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.

---

病毒扩散得很快，如今你的任务是尽量地经过安装防火墙来隔离病毒。

假设世界由二维矩阵组成，0 表示该区域未感染病毒，而 1 表示该区域已感染病毒。能够在任意 2 个四方向相邻单元之间的共享边界上安装一个防火墙（而且只有一个防火墙）。

天天晚上，病毒会从被感染区域向相邻未感染区域扩散，除非被防火墙隔离。现因为资源有限，天天你只能安装一系列防火墙来隔离其中一个被病毒感染的区域（一个区域或连续的一片区域），且该感染区域对未感染区域的威胁最大且保证惟一。

你须要努力使得最后有部分区域不被病毒感染，若是能够成功，那么返回须要使用的防火墙个数; 若是没法实现，则返回在世界被病毒所有感染时已安装的防火墙个数。 

示例 1：

输入: 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]]
输出: 10
说明:
一共有两块被病毒感染的区域: 从左往右第一块须要 5 个防火墙，同时若该区域不隔离，晚上将感染 5 个未感染区域（即被威胁的未感染区域个数为 5）;
第二块须要 4 个防火墙，同理被威胁的未感染区域个数是 4。所以，第一天先隔离左边的感染区域，通过一晚后，病毒传播后世界以下:
[[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]]
第二题，只剩下一块未隔离的被感染的连续区域，此时须要安装 5 个防火墙，且安装完毕后病毒隔离任务完成。

示例 2：

输入: grid = 
[[1,1,1],
 [1,0,1],
 [1,1,1]]
输出: 4
说明: 
此时只须要安装 4 面防火墙，就有一小区域能够幸存，不被病毒感染。
注意不须要在世界边界创建防火墙。 

示例 3:

输入: 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]]
输出: 13
说明: 
在隔离右边感染区域后，隔离左边病毒区域只须要 2 个防火墙了。 

说明:

1. grid 的行数和列数范围是 [1, 50]。
2.  grid[i][j] 只包含 0 或 1 。
3. 题目保证每次选取感染区域进行隔离时，必定存在惟一一个对未感染区域的威胁最大的区域。

---

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 }