<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:
    def shortestPathAllKeys(self, grid: List[str]) -> int:
        m, n = len(grid), len(grid[0])

        poi = { cell: (i, j)
                for i, row in enumerate(grid)
                for j, cell in enumerate(row)
                if cell not in '.#'
            }
        
        def neighbors(i, j):
            for di, dj in [[1, 0], [-1, 0], [0, 1], [0, -1]]:
                ti, tj = i+di, j+dj
                if 0 <= ti < m and 0 <= tj < n:
                    yield ti, tj
        
        # build graph for poi
        def bts_from(src):
            i, j = poi[src]
            seen = [[False]*n for _ in range(m)]
            seen[i][j] = True
            q = collections.deque([(i, j, 0)])
            poi_dist = {}
            while q:
                ci, cj, dist = q.popleft()
                if src != grid[ci][cj] != '.':
                    poi_dist[grid[ci][cj]] = dist
                    continue # stop when we meet a new poi
                for ni, nj in neighbors(ci, cj):
                    if grid[ni][nj] != '#' and not seen[ni][nj]:
                        seen[ni][nj] = True
                        q.append((ni, nj, dist+1))
            return poi_dist

        poi_dists = {place: bts_from(place) for place in poi}
        target_state = 2 ** sum(p.islower() for p in poi) - 1
        # print(poi_dists)
        # {
        #     '@': {'a': 2}, 
        #     'a': {'@': 2, 'A': 4, 'B': 4}, 
        #     'b': {'A': 2}, 
        #     'A': {'b': 2, 'B': 2, 'a': 4}, 
        #     'B': {'A': 2, 'a': 4}
        # }
        # print(target_state)

        pq = [(0, '@', 0)]
        final_dist = collections.defaultdict(lambda: float('inf'))
        final_dist[('@', 0)] = 0
        while pq:
            print(pq)
            dist, place, state = heapq.heappop(pq)
            
            if final_dist[(place, state)] < dist:
                continue
            
            # got all the keys?
            if state == target_state:
                return dist

            # try go to each available poi
            for dest, dist2 in poi_dists[place].items():
                curr_src_dest_state = state 

                if dest.islower(): # key
                    curr_src_dest_state |= (1 << (ord(dest) - ord('a'))) # update state with key
                elif dest.isupper(): # lock
                    if not (state & (1 << (ord(dest) - ord('A')))): # no key
                        continue

                total_dist = dist + dist2
                if total_dist < final_dist[(dest, curr_src_dest_state)]:
                    final_dist[(dest, curr_src_dest_state)] = total_dist
                    heapq.heappush(pq, (total_dist, dest, curr_src_dest_state))

        return -1

        #=============
        # R, C = len(grid), len(grid[0])

        # # The points of interest
        # location = {v: (r, c)
        #             for r, row in enumerate(grid)
        #             for c, v in enumerate(row)
        #             if v not in '.#'}

        # def neighbors(r, c):
        #     for cr, cc in ((r-1, c), (r, c-1), (r+1, c), (r, c+1)):
        #         if 0 <= cr < R and 0 <= cc < C:
        #             yield cr, cc

        # # The distance from source to each point of interest
        # def bfs_from(source):
        #     r, c = location[source]
        #     seen = [[False] * C for _ in range(R)]
        #     seen[r][c] = True
        #     queue = collections.deque([(r, c, 0)])
        #     dist = {}
        #     while queue:
        #         r, c, d = queue.popleft()
        #         if source != grid[r][c] != '.':
        #             dist[grid[r][c]] = d
        #             continue # Stop walking from here if we reach a point of interest
        #         for cr, cc in neighbors(r, c):
        #             if grid[cr][cc] != '#' and not seen[cr][cc]:
        #                 seen[cr][cc] = True
        #                 queue.append((cr, cc, d+1))
        #     return dist        

        # dists = {place: bfs_from(place) for place in location}
        # target_state = 2 ** sum(p.islower() for p in location) - 1
        # print(dists)
        # print(target_state)
        # #Dijkstra
        # pq = [(0, '@', 0)]
        # final_dist = collections.defaultdict(lambda: float('inf'))
        # final_dist['@', 0] = 0
        # while pq:
        #     d1, place, state = heapq.heappop(pq)
        #     if final_dist[place, state] < d1: continue
        #     if state == target_state: return d1
        #     for destination, d2 in dists[place].items():
        #         state2 = state
        #         if destination.islower(): #key
        #             state2 |= (1 << (ord(destination) - ord('a')))
        #         elif destination.isupper(): #lock
        #             if not(state & (1 << (ord(destination) - ord('A')))): #no key
        #                 continue

        #         if d1 + d2 < final_dist[destination, state2]:
        #             final_dist[destination, state2] = d1 + d2
        #             heapq.heappush(pq, (d1+d2, destination, state2))

        # return -1
        
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