<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 {
public:
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        vector<pair<int, int>> pos;
        for (int i = 0; i<matrix.size(); i++) {
            for (int j = 0; j<matrix[i].size(); j++) {
                pos.emplace_back(i, j);
            }
        }
        sort(pos.begin(), pos.end(), [&](auto x, auto y) {
            return matrix[x.first][x.second] < matrix[y.first][y.second];
        });
        vector<int> dp(pos.size(), 1);
        array<int, 4> dx{-1, 1, 0, 0};
        array<int, 4> dy{0, 0, -1, 1};
        int m = int(matrix[0].size());
        for (auto [i, j] : pos) {
            for (int k = 0; k<4; k++) {
                auto nx = i+dx[k], ny = j+dy[k];
                if (nx < 0 || ny < 0 || nx >= int(matrix.size()) || ny >= int(matrix[0].size())) continue;
                if (matrix[nx][ny] < matrix[i][j]) {
                    dp[i * m + j] = max(dp[i * m + j], dp[nx * m + ny] + 1);
                }
            }
        }
        return *max_element(dp.begin(), dp.end());
    }
};
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