<p>Given an <code>m x n</code> grid. Each cell of the grid has a sign pointing to the next cell you should visit if you are currently in this cell. The sign of <code>grid[i][j]</code> can be:</p>

<ul>
	<li><code>1</code> which means go to the cell to the right. (i.e go from <code>grid[i][j]</code> to <code>grid[i][j + 1]</code>)</li>
	<li><code>2</code> which means go to the cell to the left. (i.e go from <code>grid[i][j]</code> to <code>grid[i][j - 1]</code>)</li>
	<li><code>3</code> which means go to the lower cell. (i.e go from <code>grid[i][j]</code> to <code>grid[i + 1][j]</code>)</li>
	<li><code>4</code> which means go to the upper cell. (i.e go from <code>grid[i][j]</code> to <code>grid[i - 1][j]</code>)</li>
</ul>

<p>Notice that there could be some signs on the cells of the grid that point outside the grid.</p>

<p>You will initially start at the upper left cell <code>(0, 0)</code>. A valid path in the grid is a path that starts from the upper left cell <code>(0, 0)</code> and ends at the bottom-right cell <code>(m - 1, n - 1)</code> following the signs on the grid. The valid path does not have to be the shortest.</p>

<p>You can modify the sign on a cell with <code>cost = 1</code>. You can modify the sign on a cell <strong>one time only</strong>.</p>

<p>Return <em>the minimum cost to make the grid have at least one valid path</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/02/13/grid1.png" style="width: 400px; height: 390px;" />
<pre>
<strong>Input:</strong> grid = [[1,1,1,1],[2,2,2,2],[1,1,1,1],[2,2,2,2]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> You will start at point (0, 0).
The path to (3, 3) is as follows. (0, 0) --&gt; (0, 1) --&gt; (0, 2) --&gt; (0, 3) change the arrow to down with cost = 1 --&gt; (1, 3) --&gt; (1, 2) --&gt; (1, 1) --&gt; (1, 0) change the arrow to down with cost = 1 --&gt; (2, 0) --&gt; (2, 1) --&gt; (2, 2) --&gt; (2, 3) change the arrow to down with cost = 1 --&gt; (3, 3)
The total cost = 3.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/02/13/grid2.png" style="width: 350px; height: 341px;" />
<pre>
<strong>Input:</strong> grid = [[1,1,3],[3,2,2],[1,1,4]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> You can follow the path from (0, 0) to (2, 2).
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/02/13/grid3.png" style="width: 200px; height: 192px;" />
<pre>
<strong>Input:</strong> grid = [[1,2],[4,3]]
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 100</code></li>
	<li><code>1 &lt;= grid[i][j] &lt;= 4</code></li>
</ul>

#include <vector>
#include <deque>
#include <algorithm>
#include <climits>

using namespace std;

int speedup = []{ ios::sync_with_stdio(0); cin.tie(0); return 0; }();
unsigned char m0[] {0, 1, 4, 8, 2};
unsigned char msk[10000], tgt[10000], dd[10000];
int dirs[] {0, 1, -2, 0, -1, 0, 0, 0, 2};

class Solution {
public:
    int minCost(vector<vector<int>>& grid) {
        int w = grid[0].size(), h = grid.size(), n = w * h;
        fill(msk, msk + n, 15);

        for (int i = 0, d = n - w; i < w; ++i) {
            msk[i] &= ~2;
            msk[i + d] &= ~8;
        }

        for (int i = 0, r = w - 1; i < n; i += w) {
            msk[i] &= ~4;
            msk[i + r] &= ~1;
        }

        for (int i = 0, p = 0; i < h; ++i) {
            for (int j = 0; j < w; ++j) {
                int t = m0[grid[i][j]] & msk[p];
                msk[p] -= t;
                tgt[p++] = t;
            }
        }

        dirs[2] = -w;
        dirs[8] = w;

        fill(dd, dd + n, 255);
        --n;
        deque<int> q;
        q.push_back(0);
        dd[0] = 1;

        while (1) {
            int nd = q.front();
            q.pop_front();
            int d = dd[nd], m = msk[nd], t = tgt[nd];
            if (nd == n) return d - 1;
            if (!d) continue;
            dd[nd] = 0;

            if (t) {
                int nb = nd + dirs[t];
                if (d < dd[nb]) dd[nb] = d, q.push_front(nb);
            }

            for (int i = 1; i < 16; i *= 2) {
                if (m & i) {
                    int nb = nd + dirs[i];
                    if (d + 1 < dd[nb]) dd[nb] = d + 1, q.push_back(nb);
                }
            }
        }
    }
};
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