<p>Given an <code>m x n</code> integers <code>matrix</code>, return <em>the length of the longest increasing path in </em><code>matrix</code>.</p>

<p>From each cell, you can either move in four directions: left, right, up, or down. You <strong>may not</strong> move <strong>diagonally</strong> or move <strong>outside the boundary</strong> (i.e., wrap-around is not allowed).</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/05/grid1.jpg" style="width: 242px; height: 242px;" />
<pre>
<strong>Input:</strong> matrix = [[9,9,4],[6,6,8],[2,1,1]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The longest increasing path is <code>[1, 2, 6, 9]</code>.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/27/tmp-grid.jpg" style="width: 253px; height: 253px;" />
<pre>
<strong>Input:</strong> matrix = [[3,4,5],[3,2,6],[2,2,1]]
<strong>Output:</strong> 4
<strong>Explanation: </strong>The longest increasing path is <code>[3, 4, 5, 6]</code>. Moving diagonally is not allowed.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[1]]
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>m == matrix.length</code></li>
	<li><code>n == matrix[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 200</code></li>
	<li><code>0 &lt;= matrix[i][j] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

class Solution {
    using Node = std::pair<int, int>;
    using Nodes = std::vector<Node>;
    using NodesQueue = std::queue<Node>;
    template <typename T>
    using NodeTo = std::vector<std::vector<T>>;
    NodeTo<Nodes> revAdjList;
    NodeTo<int> outDegrees;
    NodeTo<int> maxDist;
    NodesQueue curLocalMaximas;
public:
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        int m = matrix.size(), n = matrix[0].size();
        revAdjList.resize(m, std::vector<Nodes>(n));
        outDegrees.resize(m, std::vector<int>(n, 0));
        maxDist.resize(m, std::vector<int>(n, 1));
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                auto &outDeg = outDegrees[i][j];
                if (i != 0 && matrix[i-1][j] > matrix[i][j]) {
                    revAdjList[i-1][j].push_back({i, j});
                    ++outDeg;
                }
                if (i != m-1 && matrix[i+1][j] > matrix[i][j]) {
                    revAdjList[i+1][j].push_back({i, j});
                    ++outDeg;
                }
                if (j != 0 && matrix[i][j-1] > matrix[i][j]) {
                    revAdjList[i][j-1].push_back({i, j});
                    ++outDeg;
                }
                if (j != n-1 && matrix[i][j+1] > matrix[i][j]) {
                    revAdjList[i][j+1].push_back({i, j});
                    ++outDeg;
                }
                if (outDeg == 0) {
                    curLocalMaximas.push({i, j});
                }
            }
        }
        int ans = 1;
        while (!curLocalMaximas.empty()) {
            Node curNode = curLocalMaximas.front();
            curLocalMaximas.pop();
            int curMaxDist = maxDist[curNode.first][curNode.second];
            for (auto adj: revAdjList[curNode.first][curNode.second]) {
                auto &neighbMaxDist = maxDist[adj.first][adj.second];
                neighbMaxDist = max(neighbMaxDist, 1 + curMaxDist);
                ans = max(ans, neighbMaxDist);
                auto &outDeg = outDegrees[adj.first][adj.second];
                --outDeg;
                if (outDeg == 0) {
                    curLocalMaximas.push({adj.first, adj.second});
                }
            }

        }
        return ans;
    }
};
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