841. 钥匙和房间

思路：标准图的DFS遍历方法，无需多说

class Solution {
public:
    void DFS(int r,vector<vector<int>>& rooms,vector<int>& vis){
        for(int i=0;i<rooms[r].size();i++){
            if(vis[rooms[r][i]]==0){
                vis[rooms[r][i]]=1;
                DFS(rooms[r][i],rooms,vis);
            }
        }
    }
    bool canVisitAllRooms(vector<vector<int>>& rooms) {
        int n=rooms.size();
        vector<int> vis(n,0);
        vis[0]=1;
        DFS(0,rooms,vis);
        for(int i=0;i<n;i++)if(vis[i]==0)return false;
        return true;
    }
};