<p>A game is played by a cat and a mouse named Cat and Mouse.</p>

<p>The environment is represented by a <code>grid</code> of size <code>rows x cols</code>, where each element is a wall, floor, player (Cat, Mouse), or food.</p>

<ul>
	<li>Players are represented by the characters <code>&#39;C&#39;</code>(Cat)<code>,&#39;M&#39;</code>(Mouse).</li>
	<li>Floors are represented by the character <code>&#39;.&#39;</code> and can be walked on.</li>
	<li>Walls are represented by the character <code>&#39;#&#39;</code> and cannot be walked on.</li>
	<li>Food is represented by the character <code>&#39;F&#39;</code> and can be walked on.</li>
	<li>There is only one of each character <code>&#39;C&#39;</code>, <code>&#39;M&#39;</code>, and <code>&#39;F&#39;</code> in <code>grid</code>.</li>
</ul>

<p>Mouse and Cat play according to the following rules:</p>

<ul>
	<li>Mouse <strong>moves first</strong>, then they take turns to move.</li>
	<li>During each turn, Cat and Mouse can jump in one of the four directions (left, right, up, down). They cannot jump over the wall nor outside of the <code>grid</code>.</li>
	<li><code>catJump, mouseJump</code> are the maximum lengths Cat and Mouse can jump at a time, respectively. Cat and Mouse can jump less than the maximum length.</li>
	<li>Staying in the same position is allowed.</li>
	<li>Mouse can jump over Cat.</li>
</ul>

<p>The game can end in 4 ways:</p>

<ul>
	<li>If Cat occupies the same position as Mouse, Cat wins.</li>
	<li>If Cat reaches the food first, Cat wins.</li>
	<li>If Mouse reaches the food first, Mouse wins.</li>
	<li>If Mouse cannot get to the food within 1000 turns, Cat wins.</li>
</ul>

<p>Given a <code>rows x cols</code> matrix <code>grid</code> and two integers <code>catJump</code> and <code>mouseJump</code>, return <code>true</code><em> if Mouse can win the game if both Cat and Mouse play optimally, otherwise return </em><code>false</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/12/sample_111_1955.png" style="width: 580px; height: 239px;" />
<pre>
<strong>Input:</strong> grid = [&quot;####F&quot;,&quot;#C...&quot;,&quot;M....&quot;], catJump = 1, mouseJump = 2
<strong>Output:</strong> true
<strong>Explanation:</strong> Cat cannot catch Mouse on its turn nor can it get the food before Mouse.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/12/sample_2_1955.png" style="width: 580px; height: 175px;" />
<pre>
<strong>Input:</strong> grid = [&quot;M.C...F&quot;], catJump = 1, mouseJump = 4
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> grid = [&quot;M.C...F&quot;], catJump = 1, mouseJump = 3
<strong>Output:</strong> false
</pre>

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

<ul>
	<li><code>rows == grid.length</code></li>
	<li><code>cols = grid[i].length</code></li>
	<li><code>1 &lt;= rows, cols &lt;= 8</code></li>
	<li><code>grid[i][j]</code> consist only of characters <code>&#39;C&#39;</code>, <code>&#39;M&#39;</code>, <code>&#39;F&#39;</code>, <code>&#39;.&#39;</code>, and <code>&#39;#&#39;</code>.</li>
	<li>There is only one of each character <code>&#39;C&#39;</code>, <code>&#39;M&#39;</code>, and <code>&#39;F&#39;</code> in <code>grid</code>.</li>
	<li><code>1 &lt;= catJump, mouseJump &lt;= 8</code></li>
</ul>

class Solution:
    def canMouseWin(self, grid: List[str], catJump: int, mouseJump: int) -> bool:
        dirs = [[1, 0], [-1, 0], [0, 1], [0, -1]]
        m, n = len(grid), len(grid[0])
        mouse_pos = cat_pos = None
        available = 0 # available steps for mouse and cat
        # Search the start pos of mouse and cat
        for i in range(m):
            for j in range(n):
                if grid[i][j] != '#':
                    available += 1
                if grid[i][j] == 'M':
                    mouse_pos = (i, j)
                elif grid[i][j] == 'C':
                    cat_pos = (i, j)
        
        @functools.lru_cache(None)
        def dp(turn, mouse_pos, cat_pos):
            # if turn == m * n * 2:
            # We already search the whole grid (9372 ms 74.3 MB)
            if turn == available * 2:
                # We already search the whole touchable grid (5200 ms 57.5 MB)
                return False
            if turn % 2 == 0:
                # Mouse
                i, j = mouse_pos
                for di, dj in dirs:
                    for jump in range(mouseJump + 1):
                        # Note that we want to do range(mouseJump + 1) instead of range(1, mouseJump + 1)
                        # considering the case that we can stay at the same postion for next turn.
                        new_i, new_j = i + di * jump, j + dj * jump
                        if 0 <= new_i < m and 0 <= new_j < n and grid[new_i][new_j] != '#':
                            # Valid pos
                            if dp(turn + 1, (new_i, new_j), cat_pos) or grid[new_i][new_j] == 'F':
                                return True
                        else:
                            # Stop extending the jump since we cannot go further
                            break
                return False
            else:
                # Cat
                i, j = cat_pos
                for di, dj in dirs:
                    for jump in range(catJump + 1):
                        new_i, new_j = i + di * jump, j + dj * jump
                        if 0 <= new_i < m and 0 <= new_j < n and grid[new_i][new_j] != '#':
                            if not dp(turn + 1, mouse_pos, (new_i, new_j)) or (new_i, new_j) == mouse_pos or grid[new_i][new_j] == 'F':
                            # This condition will also handle the case that the cat cannot jump through the mouse
                                return False
                        else:
                            break
                return True
                
        return dp(0, mouse_pos, cat_pos)
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