<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:
    struct node{
        int dis;
        int pos;
        bool operator < (const node& rhs) const{
            return rhs.dis<dis;
        }
    };
    priority_queue<node> pq;
    vector<bool> vis;
    vector<int> dis;
    vector<pair<int,int>> v[10000];
    void dij(int s){
        pq.push((node){0,s});
        dis[s]=0;
        while(pq.size()){
            int x=pq.top().pos;
            pq.pop();
            if(vis[x]) continue;
            vis[x]=1;
            for(auto [y,d]:v[x]){
                if(dis[y]>dis[x]+d){
                    dis[y]=dis[x]+d;
                    if(!vis[y]){
                        pq.push((node){dis[y],y});
                    }
                }
            }
        }
    }
    int minCost(vector<vector<int>>& grid) {
        vis.resize(grid.size()*grid[0].size());
        dis.resize(grid.size()*grid[0].size(),0x3f3f3f3f);
        int n=grid.size();
        int m=grid[0].size();
        for(int i=0;i<n-1;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j]==3){
                    v[(i*m)+j].push_back(make_pair(i*m+m+j,0));
                }else{
                    v[(i*m)+j].push_back(make_pair(i*m+m+j,1));
                }
            }
        }
        for(int i=1;i<n;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j]==4){
                    v[(i*m)+j].push_back(make_pair(i*m-m+j,0));
                }else{
                    v[(i*m)+j].push_back(make_pair(i*m-m+j,1));
                }
            }
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<m-1;j++){
                if(grid[i][j]==1){
                    v[(i*m)+j].push_back(make_pair(i*m+j+1,0));
                }else{
                    v[(i*m)+j].push_back(make_pair(i*m+j+1,1));
                }
            }
        }
        for(int i=0;i<n;i++){
            for(int j=1;j<m;j++){
                if(grid[i][j]==2){
                    v[(i*m)+j].push_back(make_pair(i*m+j-1,0));
                }else{
                    v[(i*m)+j].push_back(make_pair(i*m+j-1,1));
                }
            }
        }
        dij(0);
        return dis[n*m-1];
    }
};
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