<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>

#define pii pair<int, int>
#define maxn 10005
#define inf 1000000000

class Solution {
public:
    vector<pii>g[maxn];
    int dis[maxn];
    int n, m;
    bool valid(int x, int y) {
        if(x >= 0 and x < n and y >= 0 and y < m) return true;
        else return false;
    }
    void bfs(int s) {
        deque<int> q;
        for(int i=0; i<maxn; i++) dis[i] = inf;
        dis[s] = 0;
        q.push_front(s);
        while(!q.empty()){
            int u = q.front();
            q.pop_front();
            for(int i=0; i<g[u].size(); i++){
                int v = g[u][i].first;
                int w = g[u][i].second;
                if(dis[u] + w < dis[v]){
                    dis[v] = dis[u] + w;
                    if(w == 1) q.push_back(v);
                    else q.push_front(v);
                }
            }
        }
    }
    int minCost(vector<vector<int>>& grid) {
        n = grid.size(), m = grid[0].size();
        for(int i=0; i<n; i++) {
            for(int j=0; j<m; j++) {
                int u = i * m + j;
                if(valid(i+1, j)) {
                    int v = (i+1) * m + j;
                    int w;
                    if(grid[i][j] == 3) w = 0;
                    else w = 1;
                    g[u].push_back({v, w});
                }
                if(valid(i-1, j)) {
                    int v = (i-1) * m + j;
                    int w;
                    if(grid[i][j] == 4) w = 0;
                    else w = 1;
                    g[u].push_back({v, w});
                } 
                if(valid(i, j+1)) {
                    int v = i * m + (j+1);
                    int w;
                    if(grid[i][j] == 1) w = 0;
                    else w = 1;
                    g[u].push_back({v, w});
                } 
                if(valid(i, j-1)) {
                    int v = i * m + (j-1);
                    int w;
                    if(grid[i][j] == 2) w = 0;
                    else w = 1;
                    g[u].push_back({v, w});
                } 
            }
        }
        bfs(0);
        int node = (n * m) - 1;
        int ans = dis[node];
        return ans;
    }
};
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