<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>>& b) {
		int n = b.size(), m = b[0].size();
		vector<vector<int>> a(n + 2, vector<int>(m + 2)), c;
		c = a;
		vector<pair<int, int>> it;
		it.reserve(n * m);
		for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++) {
				a[i + 1][j + 1] = b[i][j];
				it.push_back({ i + 1, j + 1});
			}
		sort(it.rbegin(), it.rend(), [&](pair<int, int> l, pair<int, int> r) {return a[l.first][l.second] < a[r.first][r.second]; });
		int ans = 0;
		for (auto[i, j] : it) {
			c[i][j] = 1;
			for (int ii = -1; ii < 2; ii++)
				for (int jj = -1; jj < 2; jj++)
					if (abs(ii) != abs(jj) && a[i][j] < a[i + ii][j + jj])
						c[i][j] = max(c[i][j], c[i + ii][j + jj] + 1);
			ans = max(ans, c[i][j]);
		}
		return ans;

	}
};
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