<p>You are given an <code>m x n</code> grid <code>grid</code> where:</p>

<ul>
	<li><code>&#39;.&#39;</code> is an empty cell.</li>
	<li><code>&#39;#&#39;</code> is a wall.</li>
	<li><code>&#39;@&#39;</code> is the starting point.</li>
	<li>Lowercase letters represent keys.</li>
	<li>Uppercase letters represent locks.</li>
</ul>

<p>You start at the starting point and one move consists of walking one space in one of the four cardinal directions. You cannot walk outside the grid, or walk into a wall.</p>

<p>If you walk over a key, you can pick it up and you cannot walk over a lock unless you have its corresponding key.</p>

<p>For some <code><font face="monospace">1 &lt;= k &lt;= 6</font></code>, there is exactly one lowercase and one uppercase letter of the first <code>k</code> letters of the English alphabet in the grid. This means that there is exactly one key for each lock, and one lock for each key; and also that the letters used to represent the keys and locks were chosen in the same order as the English alphabet.</p>

<p>Return <em>the lowest number of moves to acquire all keys</em>. If it is impossible, return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/23/lc-keys2.jpg" style="width: 404px; height: 245px;" />
<pre>
<strong>Input:</strong> grid = [&quot;@.a..&quot;,&quot;###.#&quot;,&quot;b.A.B&quot;]
<strong>Output:</strong> 8
<strong>Explanation:</strong> Note that the goal is to obtain all the keys not to open all the locks.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/23/lc-key2.jpg" style="width: 404px; height: 245px;" />
<pre>
<strong>Input:</strong> grid = [&quot;@..aA&quot;,&quot;..B#.&quot;,&quot;....b&quot;]
<strong>Output:</strong> 6
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/23/lc-keys3.jpg" style="width: 244px; height: 85px;" />
<pre>
<strong>Input:</strong> grid = [&quot;@Aa&quot;]
<strong>Output:</strong> -1
</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 an English letter, <code>&#39;.&#39;</code>, <code>&#39;#&#39;</code>, or <code>&#39;@&#39;</code>.&nbsp;</li>
	<li>There is exactly one&nbsp;<code>&#39;@&#39;</code>&nbsp;in the grid.</li>
	<li>The number of keys in the grid is in the range <code>[1, 6]</code>.</li>
	<li>Each key in the grid is <strong>unique</strong>.</li>
	<li>Each key in the grid has a matching lock.</li>
</ul>

class Solution {
public:
    int shortestPathAllKeys(vector<string>& grid) {
        vector<vector<int>> del{{0,1},{1,0},{-1,0},{0,-1}};
        int n = grid.size(), m = grid[0].size();
        // {dist,x,y,mask}
        priority_queue<vector<int>> pq;
        pair<int,int> src;
        int kc=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j]=='@'){
                    src.first = i , src.second = j;
                }
                if(grid[i][j]>='a' && grid[i][j]<='z'){
                    kc++;
                }
            }
        }
        vector<vector<vector<int>>> dist(n,vector<vector<int>> (m,vector<int> (1<<kc,1e9)));
        pq.push({0,src.first,src.second,((1<<kc)-1)});
        dist[src.first][src.second][(1<<kc)-1]=0;
        while(!pq.empty()){
            auto it = pq.top();
            pq.pop();
            int x = it[1] , y = it[2];
            int mask = it[3];
            if(grid[x][y]>='a' && grid[x][y]<='z')mask=mask&(~(1<<(grid[x][y]-'a')));
            dist[x][y][mask]=-it[0];
            if(mask==0)return dist[x][y][mask];
            for(auto ng:del){
                int nx = x+ng[0],ny = y+ng[1];
                if(nx>=0 && nx<n && ny>=0 && ny<m && grid[nx][ny]!='#'){
                    if(grid[nx][ny]>='A' && grid[nx][ny]<='Z'){
                        if(((mask>>(grid[nx][ny]-'A'))&1)==0){
                            if(dist[nx][ny][mask]>dist[x][y][mask]+1){
                                dist[nx][ny][mask] = dist[x][y][mask]+1;
                                pq.push({-dist[nx][ny][mask],nx,ny,mask});
                            }
                        }
                    }else{
                        if(dist[nx][ny][mask]>dist[x][y][mask]+1){
                            dist[nx][ny][mask] = dist[x][y][mask]+1;
                            pq.push({-dist[nx][ny][mask],nx,ny,mask});
                        }
                    }
                    
                }
            }

        }
        return -1;
    }
};
54675