<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>

constexpr unsigned char K_MASK = 0x7F, K_OFFS = 0x7;
constexpr unsigned char L_MASK = 0x1F, L_OFFS = 0x5;
class Solution {
public:
    int shortestPathAllKeys(vector<string>& grid) {
        int m = grid.size(), n = grid[0].size();
        int dirs[5] = {0, 1, 0, -1, 0};
        std::queue<int> q;
        unsigned char total = 0;
        int start = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (!(grid[i][j] & grid[i][j] - 1)) start = (j << L_OFFS ^ i) << K_OFFS;
                else if (!(grid[i][j] >> L_OFFS ^ 3)) total = total << 1 ^ 1;
            }
        }
        bool vis[m * n * (total + 1)];
        std::memset(vis, false, sizeof(vis));
        q.push(start);
        int depth = 0;
        while (!q.empty()) {
            int sz = q.size();
            while (sz--) {
                int tmp = q.front(); q.pop();
                unsigned char key = tmp & K_MASK, x = tmp >> K_OFFS & L_MASK, y = tmp >> K_OFFS + L_OFFS;
                if (key == total) return depth;
                for (int i = 0; i < 4; ++i) {
                    unsigned char dx = x + dirs[i], dy = y + dirs[i + 1], dk = key, ch = 0;
                    if (dx > -1 && dx < m && dy > -1 && dy < n && grid[dx][dy] != '#') {
                        ch = grid[dx][dy];
                        if (!(ch >> L_OFFS ^ 2) && (ch & L_MASK) && !(key & 1 << (ch & L_MASK) - 1)) continue;
                        if (!(ch >> L_OFFS ^ 3)) dk |= 1 << (ch & L_MASK) - 1;
                        if (!vis[dk * m * n + dx * n + dy]) {
                            int next = dk ^ (dy << L_OFFS ^ dx) << K_OFFS;
                            q.push(next);
                            vis[dk * m * n + dx * n + dy] = true;
                        }
                    }
                }

            }
            ++depth;
        }
        return -1;
    }
};
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