[Google] LeetCode 1293 Shortest Path in a Grid with Obstacles Elimination 思维+BFS+贪心

You are given an m x n integer matrix grid where each cell is either $0$ (empty) or $1$ (obstacle). You can move up, down, left, or right from and to an empty cell in one step.

Return the minimum number of steps to walk from the upper left corner $(0, 0)$ to the lower right corner $(m - 1, n - 1)$ given that you can eliminate at most $k$ obstacles. If it is not possible to find such walk return $-1$.

Solution

需要从起点 $(0,0)$ 走到 $(m-1,n-1)$，且可以最多消去 $k$ 个障碍。我们用 $BFS$ 来解决，具体来说：由于一个位置可能有很多种访问方式，我们只需要贪心地保留所剩消除数目较多的路线即可

点击查看代码

class Solution {
private:
    int dir[4][2] = {0,1, 1,0, 0,-1, -1,0};
    
    int check(int x, int y, int r, int c){
        if(x<0||y<0||x>=r||y>=c)return 0;
        return 1;
    }
    
public:
    int shortestPath(vector<vector<int>>& grid, int k) {
       queue<vector<int>> q;
       int vis[41][41];
       int r = grid.size(), c = grid[0].size();
       for(int i=0;i<41;i++)
           for(int j=0;j<41;j++)vis[i][j]=-1;
        
        // x,y,path_length, #available remaining obstacles to remove
        q.push({0,0,0,k});
        while(!q.empty()){
            auto a = q.front();
            int x = a[0], y = a[1];
            q.pop();
            
            // arrive the target
            if(x==r-1 && y==c-1)return a[2];
            
            //if obstacle
            if(grid[x][y]){
                if(a[3]>0)a[3]--;
                else continue;
            }
            
            // visited and remaining number of obstacles is more
            if(vis[x][y]!=-1 && vis[x][y]>=a[3])continue;
            vis[x][y] = a[3];
            
            for(int i=0;i<4;i++){
                int nx = x+dir[i][0], ny = y+dir[i][1];
                if(check(nx,ny,r,c)){
                    q.push({nx,ny,a[2]+1,a[3]});
                }
            }
        }
        return -1;
    }
};