<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>

class Solution
{
public:

    enum Direction { Right = 0, Left = 1, Down = 2, Up = 3 };
    static constexpr int deltas[4][2] = { { 0, 1 }, { 0, -1 }, { 1, 0 }, { -1, 0 } };

    int minCost(std::vector<std::vector<int>> &grid)
    {
        auto compare = [] (auto &left, auto &right)
        {
            return std::get<2>(left) > std::get<2>(right);
        };
        auto m = grid.size();
        auto n = grid[0].size();
        int minCost = INT_MAX;
        typedef int Column;
        typedef int Cost;
        typedef int Row;
        typedef std::tuple<Row, Column, Cost> Triple;
        typedef std::vector<Triple> Triples;
        std::priority_queue<Triple, Triples, decltype(compare)> q(compare);
        std::vector<std::vector<int>> minCosts(m, std::vector<int>(n, INT_MAX));
        q.push({ 0, 0, 0 });
        while (!q.empty())
        {
            auto [row, column, cost] = q.top();
            q.pop();

            if (  1 + row == m
               && 1 + column == n)
            {
                minCost = std::min(minCost, cost);
                continue;
            }

            if (  row < 0
               || column < 0
               || row >= m
               || column >= n
               || cost >= minCosts[row][column])
            {
                continue;
            }
            minCosts[row][column] = cost;

            for (auto direction : { Right, Left, Up, Down })
            {
                auto newRow = row + deltas[direction][0];
                auto newColumn = column + deltas[direction][1];
                q.push({ newRow, newColumn, 1 + direction == grid[row][column] ? cost : cost + 1 });
            }

        }
        return minCost;
    }
};
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