<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>

#define ll long long
class Solution {
public:
    int maxDistance(vector<vector<int>>& grid) {
        vector<pair<ll,ll> > cur;
        ll n = grid.size();
        ll m = grid[0].size();
        ll max = 1000000007;
        vector<vector<ll> > dis(n,vector<ll>(m,max));
        for(ll i=0; i<n;i++){
            for(ll j=0; j<m; j++){
                if(grid[i][j]==1){
                    dis[i][j] = 0;
                    cur.push_back({i,j});
                }
            }
        }
        ll ans = -1;
        ll steps = 0;
        vector<ll> x = {0,0,1,-1};
        vector<ll> y = {1,-1,0,0};
        while(!cur.empty()){
            steps++;
            vector<pair<ll,ll> > next;
            for(ll i=0; i<cur.size(); i++){
                ll xx = cur[i].first;
                ll yy = cur[i].second;
                for(ll u=0; u<4; u++){
                    ll idx = xx+x[u];
                    ll idy = yy+y[u];
                    if(idx>=0 && idy>=0 && idx<n && idy<m && dis[idx][idy]==max){
                        dis[idx][idy] = 1;
                        next.push_back({idx,idy});
                    }
                }
            }
            cur.clear();
            cur = next;
            if(cur.empty()){
                break;
            }
            ans = fmax(ans,steps);
        }
        return ans;
    }
};
36656