<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:
    class Edge{
        public:
            int x, y, tox, toy, d;
            Edge(){}
            Edge(int a, int b, int c, int e, int f):x(a), y(b), tox(c), toy(e), d(f){}
    };
    class Node{
        public:
            int x,y,dist;
            Node(){}
            Node(int a, int v, int c):x(a), y(v), dist(c){}
            bool operator <(const Node& a) const{
                return dist>a.dist;
            }
    };
    int dirx[4] = {0,0,1,-1}, diry[4] = {1,-1,0,0};
    int minCost(vector<vector<int>>& grid) {
        int ans = 0;
        int n = grid.size(), m=grid[0].size();
        vector<vector<int>> dp(n, vector<int>(m,1e9));
        vector<vector<vector<Edge>>> edges(n, vector<vector<Edge>>(m, vector<Edge>(0)));
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                for(int k=0;k<4;k++){
                    int ii = i+dirx[k], jj = j+diry[k];
                    if(ii<0||jj<0||ii>=n||jj>=m) continue;
                    edges[i][j].push_back(Edge(i,j,ii,jj,k+1!=grid[i][j]));
                }
            }
        }
        priority_queue<Node>q;
        vector<vector<int>> visited(n, vector<int>(m,0));
        q.push(Node(0,0,0));
        dp[0][0] = 0;
        while(!q.empty()){
            auto now = q.top();
            q.pop();
            if(visited[now.x][now.y]) continue;
            visited[now.x][now.y] = 1;
            for(auto e: edges[now.x][now.y]){
                if(now.dist+e.d < dp[e.tox][e.toy]){
                    dp[e.tox][e.toy] = now.dist+e.d;
                    q.push(Node(e.tox, e.toy, dp[e.tox][e.toy]));
                }
            }
        }
        return dp[n-1][m-1];
    }
};
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