<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 which_way(const pair<int, int>& way){
        if(way == make_pair(0, 1)){
            return 1;
        } else if(way == make_pair(0, -1)){
            return 2;
        } else if(way == make_pair(1, 0)){
            return 3;
        } else {
            assert(way == make_pair(-1, 0));
            return 4;
        }
    }
    int minCost(vector<vector<int>>& grid) {
        int R = size(grid), C = size(grid[0]);
        const int dx[4] = {1, 0, -1, 0};
        const int dy[4] = {0, 1, 0, -1};
        auto inside = [&](int r, int c) -> bool{
            return r >= 0 && r < R && c >= 0 && c < C;
        };
        auto flatten = [&](int r, int c) -> int{
            return r * C + c;
        };
        auto unflatten = [&](int value) -> pair<int, int> {
            return make_pair(value / C, value % C);
        };
        vector<vector<pair<int, int>>> g(R * C);
        for(int r = 0; r < R; r++){
            for(int c = 0; c < C; c++){
                for(int dir = 0; dir < 4; dir++){
                    int r2 = dx[dir] + r;
                    int c2 = dy[dir] + c;
                    if(inside(r2, c2)){
                        if(which_way(make_pair(dx[dir], dy[dir])) == grid[r][c]){
                            g[flatten(r, c)].push_back({flatten(r2, c2), 0});
                        } else {
                            g[flatten(r, c)].push_back({flatten(r2, c2), 1});
                        }
                    }
                }
            }
        }
        int res = 0;
        set<pair<int, int>> st({{0, 0}});
        vector<int> cost(R * C, 1e9);
        cost[0] = 0;
        while(!st.empty()){
            auto [cur_cost, node] = *st.begin();
            st.erase(st.begin());
            for(auto& [next_node, next_cost]: g[node]){
                int new_cost = cur_cost + next_cost;
                if(new_cost < cost[next_node]){
                    cost[next_node] = new_cost;
                    st.insert({new_cost, next_node});
                }
            }
        }
        return cost[flatten(R - 1, C - 1)];
    }
};
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