<p>A game on an <strong>undirected</strong> graph is played by two players, Mouse and Cat, who alternate turns.</p>

<p>The graph is given as follows: <code>graph[a]</code> is a list of all nodes <code>b</code> such that <code>ab</code> is an edge of the graph.</p>

<p>The mouse starts at node <code>1</code> and goes first, the cat starts at node <code>2</code> and goes second, and there is a hole at node <code>0</code>.</p>

<p>During each player&#39;s turn, they <strong>must</strong> travel along one&nbsp;edge of the graph that meets where they are.&nbsp; For example, if the Mouse is at node 1, it <strong>must</strong> travel to any node in <code>graph[1]</code>.</p>

<p>Additionally, it is not allowed for the Cat to travel to the Hole (node <code>0</code>).</p>

<p>Then, the game can end in three&nbsp;ways:</p>

<ul>
	<li>If ever the Cat occupies the same node as the Mouse, the Cat wins.</li>
	<li>If ever the Mouse reaches the Hole, the Mouse wins.</li>
	<li>If ever a position is repeated (i.e., the players are in the same position as a previous turn, and&nbsp;it is the same player&#39;s turn to move), the game is a draw.</li>
</ul>

<p>Given a <code>graph</code>, and assuming both players play optimally, return</p>

<ul>
	<li><code>1</code>&nbsp;if the mouse wins the game,</li>
	<li><code>2</code>&nbsp;if the cat wins the game, or</li>
	<li><code>0</code>&nbsp;if the game is a draw.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/17/cat1.jpg" style="width: 300px; height: 300px;" />
<pre>
<strong>Input:</strong> graph = [[2,5],[3],[0,4,5],[1,4,5],[2,3],[0,2,3]]
<strong>Output:</strong> 0
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/17/cat2.jpg" style="width: 200px; height: 200px;" />
<pre>
<strong>Input:</strong> graph = [[1,3],[0],[3],[0,2]]
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>3 &lt;= graph.length &lt;= 50</code></li>
	<li><code>1&nbsp;&lt;= graph[i].length &lt; graph.length</code></li>
	<li><code>0 &lt;= graph[i][j] &lt; graph.length</code></li>
	<li><code>graph[i][j] != i</code></li>
	<li><code>graph[i]</code> is unique.</li>
	<li>The mouse and the cat can always move.&nbsp;</li>
</ul>

class Solution {
public:
    int catMouseGame(vector<vector<int>>& graph) {
        /*
            for each (step,mouse_pos,cat_pos) the state can be mouse_win, cat_win,or draw
            when mouse_pos == 0: mouse win
            when mouse_pos == cat_pos: cat win
            when step >= 2n (n nodes): draw
        */
        int n=graph.size();
        vector<vector<vector<int>>> dp(6*n, vector(n, vector(n, -1)));
        function<int(int,int,int)> search = [&](int step, int x, int y) -> int {
            if(step == 6*n) return 0; // draw
            if(x == y) return dp[step][x][y]=2; // cat win
            if(x == 0) return dp[step][x][y]=1; // mouse win
            
            if(dp[step][x][y] != -1) return dp[step][x][y];

            int turn=step%2;
            if(turn == 0) { // mouse turn
                bool hasDraw=false;
                for(int p : graph[x]) {
                    int nxt=search(step+1, p, y);
                    if(nxt == 1) return dp[step][x][y] = 1;
                    if(nxt == 0) hasDraw=true;
                }
                if(hasDraw) {
                    return dp[step][x][y] = 0;
                } else {
                    return dp[step][x][y] = 2;
                }
            } else { // cat turn
                bool hasDraw=false;
                for(int p : graph[y]) {
                    if(p==0) continue;
                    int nxt=search(step+1, x, p);
                    if(nxt == 2) return dp[step][x][y] = 2;
                    if(nxt == 0) hasDraw=true;
                }
                if(hasDraw) {
                    return dp[step][x][y] = 0;
                } else {
                    return dp[step][x][y] = 1;
                }
            }
        };
        return search(0, 1, 2);
    }
};
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