<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:
    int minCost(vector<vector<int>>& grid) {
        struct Edge {
            int cost;
            std::size_t dest;

            [[nodiscard]] bool operator<(const Edge& that) const {return this->cost < that.cost; }
            [[nodiscard]] bool operator>(const Edge& that) const {return this->cost > that.cost; }
        };

        const auto m = grid.size();
        const auto n = grid.back().size();

        auto toFlatIndex = [&] (std::size_t x, std::size_t y) -> std::size_t {
            return (x * n) + y;  
        };

        std::vector<std::vector<Edge>> graph(m * n);

        if (graph.size() == 1) {
            return 0;
        }

        for (std::size_t row_i = 0; row_i < m; ++row_i) {
            const auto& row = grid[row_i];

            for (std::size_t col_i = 0; col_i < n; ++col_i) {
                const auto& v = row[col_i];

                auto& edges = graph.at(toFlatIndex(row_i, col_i));

                if (row_i > 0) { edges.push_back(Edge{v == 4 ? 0 : 1, toFlatIndex(row_i - 1, col_i)}); }
                if (col_i > 0) { edges.push_back(Edge{v == 2 ? 0 : 1, toFlatIndex(row_i, col_i - 1)}); }

                if (row_i < m - 1) { edges.push_back(Edge{v == 3 ? 0 : 1, toFlatIndex(row_i + 1, col_i)}); }
                if (col_i < n - 1) { edges.push_back(Edge{v == 1 ? 0 : 1, toFlatIndex(row_i, col_i + 1)}); }
            }
        }

        // Now I can use djikstra to find the single-pair, shorted-path
        std::priority_queue<Edge, std::vector<Edge>, std::greater<Edge>> queue;

        for (const auto& edge : graph[0]) {
            queue.push(edge);
        }

        std::vector<bool> visited(graph.size(), false);
        visited[0] = true;

        while (!queue.empty()) {
            const auto curr = queue.top();
            queue.pop();

            if (curr.dest == graph.size() - 1) {
                return curr.cost;
            }

            if (visited[curr.dest] == false) {
                visited[curr.dest] = true;

                // Explore it if we haven't seen it before...
                for (const auto& edge : graph[curr.dest]) {
                    if (visited[edge.dest] == false) {
                        queue.push(
                            Edge {
                                edge.cost + curr.cost,
                                edge.dest
                            }
                        );
                    }
                }
            }
        }

        assert(false); // This shouldn't happen...

        return -1;
    }
};
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