<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) 
    {
        int n = grid.size();
        int m = grid[0].size();
        pair<int,int> src = {0,0};
        vector<vector<int>> dis(n, vector<int>(m, INT_MAX));
        dis[0][0] = 0;
        int src_row = 0, src_col = 0;
        typedef tuple <int,int,int> triplet; // distance, row, col
        set<triplet> s;// open set of nodes
        for (int i = 0; i < n; i ++)
        {
            for (int j = 0; j < m; j++)
            {
                if (i == 0 && j == 0)
                    s.insert({0, src_row, src_col});
                else
                    s.insert({INT_MAX, i, j});
            }
        }

        int dir[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        while (!s.empty())
        {
            auto [distance, r, c] = *s.begin();
            s.erase(s.begin());

            for (int i = 0; i < 4; i++) // all 4 directions
            {
                int nr = r + dir[i][0];
                int nc = c + dir[i][1];
                if (nr < 0 || nr >= n || nc < 0 || nc >= m) continue;

                if (grid[r][c] - 1 == i && dis[nr][nc] > distance)
                {
                    s.erase({dis[nr][nc], nr, nc});
                    dis[nr][nc] = distance;
                    s.insert({distance, nr, nc});
                }
                else if (dis[nr][nc] > distance + 1)
                {
                    s.erase({dis[nr][nc], nr, nc});
                    dis[nr][nc] = distance + 1;
                    s.insert({distance + 1, nr, nc});
                }
            }
        }

        return dis[n-1][m-1];

    }
};
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