leetcode 1020. Number of enclaves (breadth-first search)

Breadth-first search.
First mark all location access as false if the location is 1 ans++ (count the number of all 1 locations)
then mark the cell that can leave the grid boundary as true and then set the access to the grid connected to it by breadth first search to 1 then ans-- subtract it

class Solution {
    
    
public:
    bool vis[510][510];
    int xx[4]={
    
    1,-1,0,0};
    int yy[4]={
    
    0,0,1,-1};
    int numEnclaves(vector<vector<int>>& grid) {
    
    
        int ans=0;
        int m=grid.size(),n=grid[0].size();
        queue<pair<int,int>> q;

        for(int i=0;i<m;i++){
    
    
            for(int j=0;j<n;j++){
    
    
                if(grid[i][j]==1)ans++;
                vis[i][j]=false;
                if(i==0||j==0||i==m-1||j==n-1){
    
    
                    if(grid[i][j]==1){
    
    
                        q.push({
    
    i,j});
                        vis[i][j]=true;
                    }
                }
            }
        }

    while(!q.empty()){
    
    
        auto t=q.front();
        q.pop();
        int x=t.first,y=t.second;
        ans--;

        for(int k=0;k<4;k++){
    
    
            int dx=x+xx[k];
            int dy=y+yy[k];

            if(dx<0||dx>=m||dy<0||dy>=n)continue;

            if(!vis[dx][dy]&&(grid[dx][dy]==1)){
    
    
                vis[dx][dy]=true;
                q.push({
    
    dx,dy});
            }
        }
    }
    return ans;
    }
};