<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:
    using pii = pair<int, int>;
    int dx[5] = {0, 0, 0, 1, -1};
    int dy[5] = {0, 1, -1, 0, 0};

    int minCost(vector<vector<int>>& grid) {
        const int N = grid.size();
        const int M = grid[0].size();
        vector<vector<pii>> graph(N * M); // (dis, node)

        for (int i = 0; i < N; ++i) {
            for (int j = 0; j < M; ++j) {
                graph[i*M+j].emplace_back(-1, -1);
                for (int k = 1; k <= 4; ++k) {
                    int next_i = i + dx[k];
                    int next_j = j + dy[k];
                    int cost = grid[i][j] == k ? 0 : 1;
                    graph[i*M+j].emplace_back(cost, next_i*M+next_j);
                }
            }
        }

        priority_queue<pii, vector<pii>, greater<pii>> pq;
        vector<int> dis(N*M, 1000000000);
        dis[0] = 0;
        pq.emplace(0, 0);

        while (!pq.empty()) {
            pii cur = pq.top();
            pq.pop();

            if (cur.first > dis[cur.second]) {
                continue;
            }

            int i = cur.second / M;
            int j = cur.second % M;

            for (int k = 1; k <= 4; ++k) {
                int next_i = i + dx[k];
                int next_j = j + dy[k];

                if (next_i < 0 || N <= next_i || next_j < 0 || M <= next_j) {
                    continue;
                }

                if (dis[next_i * M + next_j] > dis[i * M + j] + graph[i * M + j][k].first) {
                    dis[next_i * M + next_j] = dis[i * M + j] + graph[i * M + j][k].first;
                    pq.emplace(dis[next_i * M + next_j], next_i * M + next_j);
                }
            }
        }

        // for (int i = 0; i < N; ++i) {
        //     for (int j = 0; j < M; ++j) {
        //         cout << dis[i*M+j] << " ";
        //     }
        //     cout << endl;
        // }

        return dis[N * M - 1];
    }
};
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