<p>You are given an <code>m x n</code> binary grid <code>grid</code> where <code>1</code> represents land and <code>0</code> represents water. An <strong>island</strong> is a maximal <strong>4-directionally</strong> (horizontal or vertical) connected group of <code>1</code>&#39;s.</p>

<p>The grid is said to be <strong>connected</strong> if we have <strong>exactly one island</strong>, otherwise is said <strong>disconnected</strong>.</p>

<p>In one day, we are allowed to change <strong>any </strong>single land cell <code>(1)</code> into a water cell <code>(0)</code>.</p>

<p>Return <em>the minimum number of days to disconnect the grid</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/24/land1.jpg" style="width: 500px; height: 169px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,1,0],[0,1,1,0],[0,0,0,0]]

<strong>Output:</strong> 2
<strong>Explanation:</strong> We need at least 2 days to get a disconnected grid.
Change land grid[1][1] and grid[0][2] to water and get 2 disconnected island.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/24/land2.jpg" style="width: 404px; height: 85px;" />
<pre>
<strong>Input:</strong> grid = [[1,1]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> Grid of full water is also disconnected ([[1,1]] -&gt; [[0,0]]), 0 islands.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 30</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

class Solution {
    int m, n;
public:
    int minDays(vector<vector<int>>& grid) {
        m = grid.size();
        n = grid[0].size();

        
        int count = isIsland(grid);// grid 變成2 0 
        if(count >= 2 || count == 0) return 0;

        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(grid[i][j] == 1){
                    grid[i][j] = 0;
                    int count = isIsland(grid);
                    if(count >= 2 || count == 0) return 1;
                    grid[i][j] = 1;
                }
            }
        }

        return 2;
    }
    
    int isIsland(vector<vector<int>>& grid){
        
        vector<int> dirs = {0,1,0,-1,0};
        vector<bool> visited(m*n+1);
        int count = 0;
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(grid[i][j] == 0 || visited[i*n+j]) continue;
                queue<pair<int,int>> q;
                q.push({i,j});

                while(!q.empty()){
                    int x = q.front().first;
                    int y = q.front().second;
                    q.pop();
                    if(visited[x*n+y]) continue;
                    visited[x*n+y] = 1;
                    for(int k = 1; k < dirs.size(); k++){
                        int nx = x + dirs[k-1];
                        int ny = y + dirs[k];
                        if(nx < 0 || ny < 0 || nx >= m || ny >= n || grid[nx][ny] == 0 || visited[nx*n+ny]) continue;
                        q.push({nx,ny});
                    }
                }
                count += 1;
                if(count > 1) return 2;
            }
        }
        return count;
    }
};

/*
x x x
x x
x

x x x
x x x 
x x x 

[1,1,0,1,1],
[1,1,1,1,1],
[1,1,0,1,1],
[1,1,0,1,1]]
看起來難點在於要如何判斷disconnect一個區域需要做多少次
900取2 -> 450*899 = 404550種可能
30*30遍歷一波 結果是?
*/
41260