<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) {
        int m = grid.size(), n = grid[0].size();
        vector<int>start;
        int keys = 0;
        for(int i = 0; i < m; i++) {
            for(int j = 0; j < n; j++){
                if(grid[i][j]=='@') start = {i,j,0,0};
                if(grid[i][j] >= 'a' and grid[i][j]<='f') keys++;
            }
        }

        queue<vector<int>>q;
        q.push(start);
        int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
        int visited[m][n][1<<keys][1<<keys];
        memset(visited,0,sizeof(visited));
        visited[start[0]][start[1]][0][0] = 1;
        int len = 0;
        while(!q.empty()){
            int size = q.size();
            while(size--){
                auto p = q.front();
                q.pop();
                cout << p[0] << " " <<  p[1] << " " << p[2] << " " << p[3] << endl;
                if(p[2] == (1<<keys)-1) return len;
                for(auto d: dir){
                    int x=p[0]+d[0], y= p[1]+d[1];
                    if(x>=0 and x < m and y >=0 and y < n and grid[x][y]!='#'){
                        if(grid[x][y]>='a' and grid[x][y] <='f' and !visited[x][y][p[2] | 1 << (grid[x][y]-'a')][p[3]]) {
                            visited[x][y][p[2] | 1 << (grid[x][y]-'a')][p[3]] =1;
                            q.push({x,y, p[2] | 1 << (grid[x][y]-'a'), p[3]});
                        }
                        else if(grid[x][y]>='A' and grid[x][y]<='F' and ((p[2] >> (grid[x][y]-'A')) & 1) and !visited[x][y][p[2]][p[3] | 1 << (grid[x][y]-'A')]){
                            visited[x][y][p[2]][p[3]  | 1 << (grid[x][y]-'A')] = 1;
                            q.push({x,y, p[2],p[3]  | 1 << (grid[x][y]-'A') });
                        } else if((grid[x][y]=='@' or grid[x][y]=='.') and !visited[x][y][p[2]][p[3]]){
                            visited[x][y][p[2]][p[3]] = 1;
                            q.push({x,y, p[2],p[3]});
                        }
                    }
                }
            }
            len++;
        }
        return -1;
    }
};
66596