Minimum Steps by Knight to reach target

/*
Logic: T.C: O(n*n) ie. no. of blocks of chess board
       BFS is used here.
*/

class Solution {
    public:
    int minStepToReachTarget(vector<int>&s,vector<int>&d,int n){
        s[0]--;s[1]--;d[0]--;d[1]--;

        deque<pair<pair<int,int>,int>> q;
        q.push_back(make_pair(make_pair(s[0],s[1]),0));
        vector <vector<bool>> check(n,vector <bool>(n,0));
        check[s[0]][s[1]]=true;
        int row[]={2,-2,1,1,2,-2,-1,-1};
        int col[]={1,1,2,-2,-1,-1,-2,2};
        int count=0;
        while(q.size()){
          pair<int,int> temp=q.front().first;
          int dist=q.front().second;
          q.pop_front();
          if(temp.first==d[0] && temp.second==d[1]) return dist;
          for(int i=0;i<8;i++){
            int new_x=temp.first+row[i];
            int new_y=temp.second+col[i];
              if(new_x>=0 && new_y>=0 && new_x<n && new_y<n && !check[new_x][new_y]){
                   q.push_back(make_pair(make_pair(new_x,new_y),dist+1));
                   check[new_x][new_y]=true;

              }
          }
        }
        return -1;

    }
};