<p>You are given a <strong>0-indexed</strong> 2D integer matrix <code>grid</code> of size <code>3 * 3</code>, representing the number of stones in each cell. The grid contains exactly <code>9</code> stones, and there can be <strong>multiple</strong> stones in a single cell.</p>

<p>In one move, you can move a single stone from its current cell to any other cell if the two cells share a side.</p>

<p>Return <em>the <strong>minimum number of moves</strong> required to place one stone in each cell</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/08/23/example1-3.svg" style="width: 401px; height: 281px;" />
<pre>
<strong>Input:</strong> grid = [[1,1,0],[1,1,1],[1,2,1]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> One possible sequence of moves to place one stone in each cell is: 
1- Move one stone from cell (2,1) to cell (2,2).
2- Move one stone from cell (2,2) to cell (1,2).
3- Move one stone from cell (1,2) to cell (0,2).
In total, it takes 3 moves to place one stone in each cell of the grid.
It can be shown that 3 is the minimum number of moves required to place one stone in each cell.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/08/23/example2-2.svg" style="width: 401px; height: 281px;" />
<pre>
<strong>Input:</strong> grid = [[1,3,0],[1,0,0],[1,0,3]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> One possible sequence of moves to place one stone in each cell is:
1- Move one stone from cell (0,1) to cell (0,2).
2- Move one stone from cell (0,1) to cell (1,1).
3- Move one stone from cell (2,2) to cell (1,2).
4- Move one stone from cell (2,2) to cell (2,1).
In total, it takes 4 moves to place one stone in each cell of the grid.
It can be shown that 4 is the minimum number of moves required to place one stone in each cell.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>grid.length == grid[i].length == 3</code></li>
	<li><code>0 &lt;= grid[i][j] &lt;= 9</code></li>
	<li>Sum of <code>grid</code> is equal to <code>9</code>.</li>
</ul>

class Solution:
    def minimumMoves(self, grid: List[List[int]]) -> int:

        # 当成1D的 回溯来做, 1D的 i 》2D的 x=i//3, y=i%3
        dp = {}
        def dfs(i, state):
            if i == 9:
                print(grid)
                for a in range(3):
                    for b in range(3):

                        if grid[a][b] != 1:
                            return float('inf')
                return 0
            if (i, state) in dp:
                return dp[(i,state)]
            res = float('inf')
            x, y = i // 3, i % 3
            if grid[x][y] > 1: # 要找一个空的地方 填一个1进去
                for a in range(3):
                    for b in range(3):
                        if grid[a][b] == 0:
                            grid[x][y] -= 1
                            grid[a][b] += 1
                            newState = ";".join("".join(str(row)) for row in grid)
                            step = abs(a-x)+abs(b-y)
                            res = min(res, dfs(i, newState) + step) # 这里要回到i, 因为不能保证1次操作就足够了
                            grid[a][b] -= 1
                            grid[x][y] += 1
            else:
                res = min(res, dfs(i+1,state) )
            dp[(i, state)] = res
            return res

        initial = ";".join(",".join(str(row)) for row in grid)
        res = dfs(0, initial)
        return res if res != float('inf') else -1
17300