<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 dp[1001][1001];
    vector<vector<int>>v={{0,0},{0,1},{0,-1},{1,0},{-1,0}};
     bool check(int row,int col,int n,int m){
        if(row<0 or row>=n or col<0 or col>=m) return 0;
        return 1;
     }
    int minCost(vector<vector<int>>& grid) {
        int n=grid.size();
        int m=grid[0].size();
        priority_queue<vector<int>,vector<vector<int>>,greater<vector<int>>>pq;
        pq.push({0,0,0});
         for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
               dp[i][j]=1e9;
            }
        }
        //  for(int i=0;i<n;i++){
        //     for(int j=0;j<m;j++){
        //         cout<<dp[i][j]<<" ";
        //     }cout<<endl;
        // }
        dp[0][0]=0;
        while(pq.size()>0){
            auto it=pq.top();
            pq.pop();
            int r=it[1],c=it[2],wt=it[0];
            if(r==n-1 and c==m-1) return wt;
            for(int i=1;i<5;i++){
                int row=r+v[i][0];
                int col=c+v[i][1];
                if(check(row,col,n,m)){
                    cout<<row<<" "<<col<<endl;
                    if(grid[r][c]==i and dp[row][col]>dp[r][c]){
                        dp[row][col]=dp[r][c];
                        pq.push({dp[row][col],row,col});
                    }
                    if(grid[r][c]!=i  and dp[row][col]>dp[r][c]+1){
                         dp[row][col]=dp[r][c]+1;
                         pq.push({dp[row][col],row,col});
                    }
                }
            }
        }
        // for(int i=0;i<n;i++){
        //     for(int j=0;j<m;j++){
        //         cout<<dp[i][j]<<" ";
        //     }cout<<endl;
        // }
        return dp[n-1][m-1];

    }
};
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