<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) {
        int n = graph.size();
        const int MOUSE = 1;
        const int CAT = 2;
        vector result(n + 2, vector(n + 2, vector(3, 0)));
        vector degree(n + 2, vector(n + 2, vector(3, vector(3, 0))));
        queue <tuple<int,int,int>> q;
                
        for (int t = 1; t < 3; t ++){
            for (int i = 1; i < n; i++) {
                // mouse win
                result[0][i][t] = MOUSE;
                q.push({0,i,t});
                
                // cat win
                result[i][i][t] = CAT;
                q.push({i,i,t});
            }
        }

        while(!q.empty()) {
            auto item = q.front();
            q.pop();
            int m, c, t;
            tie(m, c, t) = item;
            int r = result[m][c][t];
            for(auto v : graph[t == MOUSE ? c : m]) {
                int m2 = t == MOUSE ? m : v, c2 = t == MOUSE ? v : c, t2 = 3 - t;

                if(c2 == 0)
                    continue;

                int r2 = result[m2][c2][t2];
                if(r2 > 0)
                    continue;

                // populate prestate
                
                // can always win
                if(r == t2) {
                    result[m2][c2][t2] = t2;
                    q.push({m2, c2, t2});
                    continue;
                }

                bool allNextMovesFailed = true;

                for(auto u : graph[t2 == CAT ? c2 : m2]) {
                    int m3 = t2 == CAT ? m2 : u, c3 = t2 == CAT ? u : c2, t3 = 3 - t2;

                    if(c3 == 0)
                        continue;

                    int r3 = result[m3][c3][t3];
                    if(r3 != t3){
                        allNextMovesFailed = false;
                        break;
                    }
                }

                if (allNextMovesFailed) {
                    result[m2][c2][t2] = 3 - t2;
                    q.push({m2, c2, t2});
                }
            }
            

        }

        return result[1][2][1];
    }
};
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