<p>Given an <code>n x n</code> <code>grid</code>&nbsp;containing only values <code>0</code> and <code>1</code>, where&nbsp;<code>0</code> represents water&nbsp;and <code>1</code> represents land, find a water cell such that its distance to the nearest land cell is maximized, and return the distance.&nbsp;If no land or water exists in the grid, return <code>-1</code>.</p>

<p>The distance used in this problem is the Manhattan distance:&nbsp;the distance between two cells <code>(x0, y0)</code> and <code>(x1, y1)</code> is <code>|x0 - x1| + |y0 - y1|</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/05/03/1336_ex1.JPG" style="width: 185px; height: 87px;" />
<pre>
<strong>Input:</strong> grid = [[1,0,1],[0,0,0],[1,0,1]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The cell (1, 1) is as far as possible from all the land with distance 2.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/05/03/1336_ex2.JPG" style="width: 184px; height: 87px;" />
<pre>
<strong>Input:</strong> grid = [[1,0,0],[0,0,0],[0,0,0]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The cell (2, 2) is as far as possible from all the land with distance 4.
</pre>

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

<ul>
	<li><code>n == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= n&nbsp;&lt;= 100</code></li>
	<li><code>grid[i][j]</code>&nbsp;is <code>0</code> or <code>1</code></li>
</ul>

class Solution {
public:
int dx[4]={-1,0,1,0};
int dy[4]={0,1,0,-1};
int valid(int x,int y,int n,int m){
    if(x<0||x>=n||y<0||y>=m)return 0;
    return 1;
}
    int maxDistance(vector<vector<int>>& v) {
        int n=v.size(),m=v[0].size();
        queue<vector<int>>q;
        vector<vector<int>>dis(n+5,vector<int>(m+5,1e9));
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(v[i][j]==0)continue;
                dis[i][j]=0;
                q.push({i,j});
            }
        }
        int ans=-1;
        while(q.size()){
            int sz=q.size();
            for(int i=0;i<sz;i++){
                auto p=q.front();q.pop();
                int x=p[0],y=p[1];
                for(int dir=0;dir<4;dir++){
                    int xx=x+dx[dir],yy=y+dy[dir];
                    if(valid(xx,yy,n,m)&&dis[xx][yy]>dis[x][y]+1){
                        dis[xx][yy]=dis[x][y]+1;
                        q.push({xx,yy});
                    }
                }
            }
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(v[i][j]!=0)continue;
                ans=max(ans,dis[i][j]);
            }
        }
        return ans==1e9?-1:ans;
    }
};
53193