<p>Given an <code>m x n</code> binary matrix <code>mat</code>, return <em>the distance of the nearest </em><code>0</code><em> for each cell</em>.</p>

<p>The distance between two adjacent cells is <code>1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/24/01-1-grid.jpg" style="width: 253px; height: 253px;" />
<pre>
<strong>Input:</strong> mat = [[0,0,0],[0,1,0],[0,0,0]]
<strong>Output:</strong> [[0,0,0],[0,1,0],[0,0,0]]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/24/01-2-grid.jpg" style="width: 253px; height: 253px;" />
<pre>
<strong>Input:</strong> mat = [[0,0,0],[0,1,0],[1,1,1]]
<strong>Output:</strong> [[0,0,0],[0,1,0],[1,2,1]]
</pre>

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

<ul>
	<li><code>m == mat.length</code></li>
	<li><code>n == mat[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>4</sup></code></li>
	<li><code>mat[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
	<li>There is at least one <code>0</code> in <code>mat</code>.</li>
</ul>

class Solution {
public:
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        int rows = mat.size(), cols = mat[0].size();
        vector<vector<int>> dist(rows, vector<int>(cols));
        int vis = 0;
        queue<pair<int,int>> q;
        int level = 0;
        for(int i=0;i<rows;++i){
            for(int j = 0;j<cols;++j){
                if(mat[i][j] == 0){
                    ++vis;
                    q.push({i,j});
                    dist[i][j] = 0;
                }
            }
        }
        int tot = rows*cols;
        while(!q.empty() && vis<tot){
            int sz = q.size();
            while(sz-- && vis<tot){
                pair<int,int> cur = q.front();
                q.pop();
                
                int i = cur.first, j = cur.second;
                vector<pair<int,int>> adjs = {{i+1,j},{i,j+1},{i-1,j},{i,j-1}};
                for(auto ad: adjs){
                    if(ad.first>=0&&ad.second>=0&&ad.first<rows&&ad.second<cols
                    && mat[ad.first][ ad.second] == 1)
                    {
                        ++vis;
                        mat[ad.first][ ad.second] = -1;
                        q.push({ad.first, ad.second});
                        dist[ad.first][ad.second] = level + 1;
                    }
                }
            }
            ++level;
        }
        return dist;
    }
};
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