<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:
    public:
    int catMouseGame(vector<vector<int>>& graph) {
        int N = graph.size();
        const int DRAW = 0, MOUSE_WIN = 1, CAT_WIN = 2;
        vector<vector<vector<int>>> result(N, vector<vector<int>>(N, vector<int>(2, DRAW)));
        vector<vector<vector<int>>> degrees(N, vector<vector<int>>(N, vector<int>(2, 0)));
        queue<tuple<int, int, int>> q;
        
        // Initialize degrees and results
        for (int m = 0; m < N; ++m) {
            for (int c = 0; c < N; ++c) {
                degrees[m][c][0] = graph[m].size();
                degrees[m][c][1] = graph[c].size();
                // Remove moves where cat moves to node 0
                degrees[m][c][1] -= count(graph[c].begin(), graph[c].end(), 0);
            }
        }
        
        // Initialize terminal states
        for (int i = 0; i < N; ++i) {
            for (int t = 0; t < 2; ++t) {
                if (i != 0) {
                    result[0][i][t] = MOUSE_WIN;
                    q.push({0, i, t});
                    result[i][i][t] = CAT_WIN;
                    q.push({i, i, t});
                }
            }
        }
        
        // BFS to determine the outcome of each state
        while (!q.empty()) {
            auto [mouse, cat, turn] = q.front();
            q.pop();
            int res = result[mouse][cat][turn];
            int prev_turn = 1 - turn;
            vector<pair<int, int>> prevStates;
            
            if (turn == 0) { // Mouse moved to this state
                for (int prev_cat : graph[cat]) {
                    if (prev_cat == 0) continue; // Cat cannot move to hole
                    prevStates.emplace_back(mouse, prev_cat);
                }
            } else { // Cat moved to this state
                for (int prev_mouse : graph[mouse]) {
                    prevStates.emplace_back(prev_mouse, cat);
                }
            }
            
            for (auto [prev_mouse, prev_cat] : prevStates) {
                if (result[prev_mouse][prev_cat][prev_turn] != DRAW) continue;
                bool canWin = (res == (prev_turn == 0 ? MOUSE_WIN : CAT_WIN));
                if (canWin) {
                    result[prev_mouse][prev_cat][prev_turn] = res;
                    q.push({prev_mouse, prev_cat, prev_turn});
                } else {
                    degrees[prev_mouse][prev_cat][prev_turn]--;
                    if (degrees[prev_mouse][prev_cat][prev_turn] == 0) {
                        int loseResult = prev_turn == 0 ? CAT_WIN : MOUSE_WIN;
                        result[prev_mouse][prev_cat][prev_turn] = loseResult;
                        q.push({prev_mouse, prev_cat, prev_turn});
                    }
                }
            }
        }
        
        return result[1][2][0];
    }

};
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