Snake and Ladder Problem

/*
T.C O(N) ie. no. of blocks
Logic:
->We will use bfs here.snl ie. snake and ladder map will keep track of
  snake and ladder from any block.
->visited array will keep track of visited block.
->queue will keep pairs where first will be block number and second will
  be no. of moves to go there.
->we have option to select 1 to 6 as we have control of dice so we
  iterated from current position upto 6 moves.
*/


class Solution{
public:
    int minThrow(int n, int arr[]){
        bool visited[30+1];
        memset(visited,false,sizeof(visited));
        map<int,int> snl;
        for(int i=0;i<2*n;i+=2){
            snl[arr[i]]=arr[i+1];
        }
        deque<pair<int,int>> q;
        q.push_back(make_pair(1,0));
        visited[1]=true;
        while(q.size()){
            pair<int,int> temp=q.front();
            q.pop_front();
            int pos=temp.first;
            int dist=temp.second;
            if(pos==30) return dist;
            for(int i=pos+1;i<=pos+6;i++){
                if(visited[i]==false){
                    if(snl[i]){
                        q.push_back(make_pair(snl[i],dist+1));
                        visited[i]=true;
                    }
                    else{
                        q.push_back(make_pair(i,dist+1));
                        visited[i]=true;
                    }
                }
            }

        }
        return 0;
    }
};