BFS on grid-CSES-labyrinth

You are given a map of a labyrinth, and your task is to find a path from start to end. You can walk left, right, up and down.

Input

The first input line has two integers n and m: the height and width of the map.


First print "YES", if there is a path, and "NO" otherwise.

If there is a path, print the length of the SHORTESST  path and its description as a 
string consisting of characters L (left), R (right), U (up), and D (down). 
You can print any valid solution.

Input:
5 8
########
#.A#...#
#.##.#B#
#......#
########


Output:
YES
9
LDDRRRRRU


//bfs on grid

bool isValid(int x,int y){

if(x<0 || x>N || y<0 || y>M )return false;

if(grid[x][y]=='#' || vis[x][y]==true) return false;

return true;
}



void bfs(int strtX,int strtY,int destX,int destY){

queue<pair<int,int>> q;
q.push({strtX,strtY});
vis[strtX][strtY]=true;
dis[strtX][strtY]=0;

int dX={-1,1,0,0};
int dY={0,0,-1,1};

while(!q.empty()){

int currX=q.front().first;
int currY=q.front().second;
q.pop();


for(int i=0;i<4;i++){
if(isValid(currX+dX[i],currY+dY[i]))
{
int newX=currX+dX[i];
int newY=currY+dY[i];

vis[newX][newY]=true
q.push({newX,newY});
dis[newX][newY]=dis[currX][currY]+1;

}
}
}

}