<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>

using T3I = tuple<int, int, int>;
class Solution {
    int memo[51][51][3];
public:
    int catMouseGame(vector<vector<int>>& graph) {
        /*
            hole: node 0
            mouse: start at node 1
            cat: start at node 2

            status 0 = draw
            status 1 = mouse win
            status 2 = cat win
        */
        int n = graph.size();
        queue<T3I> q;
        for(int c=0; c<n; c++){
            memo[0][c][1] = 1;
            memo[0][c][2] = 1;
            q.push({0, c, 1});
            q.push({0, c, 2});
        }

        for(int m=1; m<n; m++){
            memo[m][m][1] = 2;
            memo[m][m][2] = 2;
            q.push({m, m, 1});
            q.push({m, m, 2});
        }

        while(!q.empty()){
            auto [m, c, turn] = q.front(); q.pop();
            int status = memo[m][c][turn];

            for(auto [m2, c2, turn2]: findAllParents(graph, m, c, turn)){
                if(memo[m2][c2][turn2]!=0) continue;

                if(turn2==status){
                    memo[m2][c2][turn2] = status;
                    q.push({m2, c2, turn2});
                }
                else if(allChildrenWin(graph, m2, c2, turn2)){
                    memo[m2][c2][turn2] = (turn2==1) ?2 :1;
                    q.push({m2, c2, turn2});
                }
            }
        }
        return memo[1][2][1];
    }

    vector<T3I> findAllParents(const vector<vector<int>>& graph, int m, int c, int turn){
        vector<T3I> neighbors;
        if(turn==1){
            for(int c_next: graph[c]){
                if(c_next!=0) neighbors.push_back({m, c_next, 2});
            }
        }
        else{
            for(int m_next: graph[m]){
                neighbors.push_back({m_next, c, 1});
            }
        }
        return neighbors;
    }

    bool allChildrenWin(vector<vector<int>>& graph, int m, int c, int turn){
        if(turn==1){
            for(int m_next: graph[m]){
                if(memo[m_next][c][2]!=2) return false;
            }
        }
        else if(turn==2){
            for(int c_next: graph[c]){
                if(c_next!=0 && memo[m][c_next][1]!=1) return false;
            }
        }
        return true;
    }
};
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