<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>

const int HOLE = 0;
const int MOUSE_START = 1;
const int CAT_START = 2;
const int MOUSE_TURN = 0;
const int CAT_TURN = 1;
const int MOUSE_WIN = 1;
const int CAT_WIN = 2;
const int TIE = 0;

class Solution {
public:
    int catMouseGame(vector<vector<int>>& graph) {
        int n = graph.size();
        int res[n][n][2];
        int degree[n][n][2];
        memset(res, 0, sizeof res);
        memset(degree, 0, sizeof degree);
        for (int i = 0; i < n; ++i) {
            for (int j = 1; j < n; ++j) {
                degree[i][j][MOUSE_TURN] = graph[i].size();
                degree[i][j][CAT_TURN] = graph[j].size();
            }
            for (int j : graph[HOLE]) {
                --degree[i][j][CAT_TURN];
            }
        }
        auto getPrevStates = [&](int m, int c, int t) {
            int pt = t ^ 1;
            vector<tuple<int, int, int>> pre;
            if (pt == CAT_TURN) {
                for (int pc : graph[c]) {
                    if (pc != HOLE) {
                        pre.emplace_back(m, pc, pt);
                    }
                }
            } else {
                for (int pm : graph[m]) {
                    pre.emplace_back(pm, c, pt);
                }
            }
            return pre;
        };
        queue<tuple<int, int, int>> q;
        for (int j = 1; j < n; ++j) {
            res[0][j][MOUSE_TURN] = res[0][j][CAT_TURN] = MOUSE_WIN;
            q.emplace(0, j, MOUSE_TURN);
            q.emplace(0, j, CAT_TURN);
        }
        for (int i = 1; i < n; ++i) {
            res[i][i][MOUSE_TURN] = res[i][i][CAT_TURN] = CAT_WIN;
            q.emplace(i, i, MOUSE_TURN);
            q.emplace(i, i, CAT_TURN);
        }
        while (!q.empty()) {
            auto [m, c, t] = q.front();
            q.pop();
            int x = res[m][c][t];
            for (auto [pm, pc, pt] : getPrevStates(m, c, t)) {
                if (res[pm][pc][pt] == TIE) {
                    bool win = (x == MOUSE_WIN && pt == MOUSE_TURN) || (x == CAT_WIN && pt == CAT_TURN);
                    if (win) {
                        res[pm][pc][pt] = x;
                        q.emplace(pm, pc, pt);
                    } else {
                        if (--degree[pm][pc][pt] == 0) {
                            res[pm][pc][pt] = x;
                            q.emplace(pm, pc, pt);
                        }
                    }
                }
            }
        }
        return res[MOUSE_START][CAT_START][MOUSE_TURN];
    }
};
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