[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，现在你的任务是尽可能地通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。可以在任意 2 个四方向相邻单元之间的共享边界上安装一个防火墙（并且只有一个防火墙）。

每天晚上，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。现由于资源有限，每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唯一。

你需要努力使得最后有部分区域不被病毒感染，如果可以成功，那么返回需要使用的防火墙个数; 如果无法实现，则返回在世界被病毒全部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: 从左往右第一块需要 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）;
第二块需要 4 个防火墙，同理被威胁的未感染区域个数是 4。因此，第一天先隔离左边的感染区域，经过一晚后，病毒传播后世界如下:
[[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，此时需要安装 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: 
此时只需要安装 4 面防火墙，就有一小区域可以幸存，不被病毒感染。
注意不需要在世界边界建立防火墙。 

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需要 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，一定存在唯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

class Solution {
    func containVirus(_ grid: [[Int]]) -> Int {
        var grid = grid
        var res:Int = 0
        var m:Int = grid.count
        var n:Int = grid[0].count
        var dirs:[[Int]] = [[-1,0],[0,1],[1,0],[0,-1]]
        while (true )
        {
            var visited:Set<Int> = Set<Int>()
            var all:[[[Int]]] = [[[Int]]]()
            for i in 0..<m
            {
                for j in 0..<n
                {
                    if grid[i][j] == 1 && !visited.contains(i * n + j)
                    {
                        var q:[Int] = [i * n + j]
                        var virus:[Int] = [i * n + j]
                        var walls:[Int] = [Int]()
                        visited.insert(i * n + j)
                        while(!q.isEmpty)
                        {
                            var t:Int = q.removeFirst()                            
                            for dir in dirs
                            {
                                var x:Int = (t / n) + dir[0]
                                var y:Int = (t % n) + dir[1]
                                if x < 0 || x >= m || y < 0 || y >= n || visited.contains(x * n + y)
                                {
                                    continue
                                }
                                if grid[x][y] == -1
                                {
                                    continue
                                }
                                else if grid[x][y] == 0
                                {
                                    walls.append(x * n + y)
                                }
                                else if grid[x][y] == 1
                                {
                                    visited.insert(x * n + y)
                                    virus.append(x * n + y)
                                    q.append(x * n + y)
                                }
                            }                            
                        }   
                        var s:Set<Int> = Set(walls)
                        var cells:[Int] = [s.count]
                        all.append([cells ,walls, virus])
                    }
                }
            }
            if all.isEmpty {break}
            all.sort(by:{(a:[[Int]],b:[[Int]]) -> Bool in 
                        return a[0][0] > b[0][0]})
            for i in 0..<all.count
            {
                if i == 0
                {
                    var virus:[Int] = all[0][2]
                    for idx in virus
                    {
                        grid[idx / n][idx % n] = -1
                    }
                    res += all[0][1].count
                }
                else
                {
                    var wall:[Int] = all[i][1]
                    for idx in wall
                    {
                        grid[idx / n][idx % n] = 1
                    }
                }
            }     
        }
        return res
    }
}