<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();
        vector<pair<int,int>>adj[n*m];
        int dr[]={-1,0,1,0};
        int dc[]={0,1,0,-1};
        for(int i=0;i<n;i++)
        {
           for(int j=0;j<m;j++)
           {
             int u=i*m+j;
             for(int k=0;k<4;k++)
             {
                int nr=i+dr[k];
                int nc=j+dc[k];
                if(!(nr>=0&&nr<n&&nc>=0&&nc<m)) continue;
                int v=nr*m+nc;
                if(grid[i][j]==1&&k==1)
                {
                    adj[u].push_back({v,0});
                    continue;
                } 
                if(grid[i][j]==2&&k==3)
                {
                    adj[u].push_back({v,0});
                    continue;
                }
                if(grid[i][j]==3&&k==2)
                {
                    adj[u].push_back({v,0});
                    continue;
                }
                if(grid[i][j]==4&&k==0)
                {
                    adj[u].push_back({v,0});
                    continue;
                }
                adj[u].push_back({v,1});
             }
           }
        }


        vector<int>dp(n*m,INT_MAX);
        dp[0]=0;
        priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>>pq;
        pq.push({0,0});
        while(!pq.empty())
        {
            auto it=pq.top();
            pq.pop();
            int dist=it.first;
            int u=it.second;
            for(auto it2:adj[u])
            {
                int v=it2.first;
                int wt=it2.second;
                int newdist=dist+wt;
                if(dp[v]>newdist)
                {
                    dp[v]=newdist;
                    pq.push({dp[v],v});
                }
            }
        }
        return dp[n*m-1];
    }
};
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