<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>

// 1 : right
// 2 : left
// 3 : down
// 4 : up
class Solution {
public:
    void print_graph(vector<list<pair<int,int>>> &graph){
        for(int i=0;i<graph.size();i++){
            cout<<i;
            for(auto neighbor : graph[i]){
                cout<<" -> "<<neighbor.first<<','<<neighbor.second;
            }
            cout<<endl;
        }
        return ;
    }
    vector<list<pair<int,int>>> prepare_graph(vector<vector<int>> &grid){
        int n=grid.size(),m=grid[0].size();
        vector<list<pair<int,int>>> graph(n*m,list<pair<int,int>>());
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                // vertex = i*m+j;
                int from=i*m+j,to,dir=grid[i][j];
                to=from+1;
                if(to<n*m && to/m==from/m){
                    if(dir==1) graph[from].push_back({to,0});
                    else graph[from].push_back({to,1});
                }
                to=from-1;
                if(to>=0 && to/m==from/m){
                    if(dir==2) graph[from].push_back({to,0});
                    else graph[from].push_back({to,1});
                }
                to=from+m;
                if(to<m*n){
                    if(dir==3) graph[from].push_back({to,0});
                    else graph[from].push_back({to,1});
                }
                to=from-m;
                if(to>=0){
                    if(dir==4) graph[from].push_back({to,0});
                    else graph[from].push_back({to,1});
                }
            }
        }
        return graph;
    }
    int minCost(vector<vector<int>>& grid) {
        int n=grid.size(),m=grid[0].size();
        vector<list<pair<int,int>>> graph = prepare_graph(grid);
        //print_graph(graph);
        //since edge weights are either 0 or 1 so we go for 0-1-BFS
        deque<pair<int,int>> dq;
        vector<bool> vis(n*m,false);
        int ans=0;
        dq.push_front({0,0});
        while(dq.size()){
            auto p=dq.front();
            dq.pop_front();
            if(!vis[p.first]){
                vis[p.first]=true;
                if(p.first==n*m-1) return p.second;
                for(auto nei : graph[p.first]){
                    if(!vis[nei.first]){
                        if(nei.second==0) dq.push_front({nei.first,p.second});
                        else dq.push_back({nei.first,p.second+1});
                    }
                }
            }
        }
        return -1;
    }
};
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