工学
单选题Spaning Tree协议（）。A 被交换机来实现虚网划分B 被路由器来划分子网C 被路由器来克服报文包循环D 被交换机来克服报文包循环
题目
被交换机来实现虚网划分
被路由器来划分子网
被路由器来克服报文包循环
被交换机来克服报文包循环
参考答案和解析
相似问题和答案
第1题：
● 开放最短路径优先协议（OSPF）采用 （22） 算法计算最佳路由。
（22）
A. Dynamic-Search
B. Bellman-Ford
C. Dijkstra
D. Spanning-Tree
第2题：
It is believed that the tradition of having a pine tree as a Christmas Tree comes from Germany.()
第3题：
IEEE802.3以太网采用_____协议以支持总线性结构，它的全称是()。
A.STP，Spanning Tree Protocol
B.STP，Shortest Tree Protocol
C.CSMA/CD Carrier Sense Multicast Access/Collision Detection
D.CSMA/CD Carrier Sent Multicast Access/CRC Detection
E.CSMA/CD Carrier Service Multiple Access/Collision Detection
F.CSMA/CD Carrier Sense Multiple Access/Collision Detection
第4题：
Refer to the exhibit. All multilayer switches are running PIM dense mode. Recipient A and Recipient B are sending IGMPv2 joi n messages to their respective multilayer switches. Which statement is true?（）
A. The Multicast Server is the root of the multicast tree. Switches 1, 3, 4, and 5 will participate in the multicast tree once pruning has taken place.
B. The Multicast Serv er is the root of the multicast tree. Switches 4 and 5 will participate in the multicast tree once pruning has taken place.
C. Switch 3 is the root of the multicast tree. Switches 3,4, and 5 will participate in the multicast tree once pruning has taken place.
D. Switch 3 is the root of the multicast tree. Switches 1,3,4, and 5 will participate in the multicast tree once pruning has taken place.
E. Switch 1 is the root of the multicast tree. Switches 1,4, and 5 will participate in the multicast tree once pruning has taken place.
F. Switch 1 is the root of the multicast tree. Switches 1,3,4, and 5 will participate in the multicast tree once pruning has taken place.
第5题：
Traversing a binary tree in preorder is equivalent to(68).
A．Traversing the forest corresponding to the binary tree in root-first order
B．Traversing the forest corresponding to the binary tree in root-last order
C．Traversing the forest corresponding to the binary tree in breadth-first order
D．None of the above
解析：前序遍历一个二叉树等价于按从树的根部、右子树、右子树查找顺序查找树。
第6题：
开放最短路径优先协议(OSPF)采用(22)算法计算最佳路由。
A．Dynamic-Search
B．Bellman-Ford
C．Dijkstra
D．Spanning-Tree
解析：OSPF(RFC2328，1998)是一种链路状态协议，这种协议要求路由器掌握完整的网络拓扑结构，并据此计算出到达目标的最佳路由。OSPF路由器通过向邻居发送一系列数据库描述分组来传送自己的数据库内容。数据库描述分组中包含了一组链路状态公告，每个链路状态公告都描述了一条链路的状态：端口的标识和连接的目标地址。发送和接收数据库描述分组的过程叫做“数据库交换过程”。当数据库交换过程结束时，路由器之间就形成了“邻接”关系。路由更新报文在邻接的路由器之间交换，当网络拓扑发生变化时，数据库的内容随之改变。路由器利用链路状态数据库存储的信息构造有向图，并通过Dijkstra的最短通路优先算法(Shortest Path First，SPF)计算最小生成树，建立和更新自己的路由表。
第7题：
A.Switch(config)#spanning-tree mode rapid-pvst
B.Switch#spanning-tree mode rapid-pvst
C.Switch(config)#spanning-tree mode 802.1w
D.Switch#spanning-tree mode 802.1w
第8题：
●Traversing a binary tree in preorder is equivalent to (68) ．
(68) A．Traversing the forest corresponding to the binary tree in root-first order
B．Traversing the forest corresponding to the binary tree in root-last order
C．Traversing the forest corresponding to the binary tree in breadth-first order
D．None of the above
【解析】前序遍历一个二叉树等价于按从树的根部、右子树、右子树查找顺序查找树。
第9题：
考虑如下代码：
class Tree{}
class Pine extends Tree{}
class Oak extends Tree{}
public class Forest {
public static void main( String[] args ) {
Tree tree = new Pine();
if( tree instanceof Pine )
System.out.println( "Pine" );
if( tree instanceof Tree )
System.out.println( "Tree" );
if( tree instanceof Oak )
System.out.println( "Oak" );
else
System.out.println( "Oops" );
}
}
则输出结果中有哪些?
A.Pine B.Tree C.Forest D.Oops E.无输出
第10题：
A. spanning-tree backbonefast
B. spanning-tree uplinkfast
C. spanning-tree portfast
D. spanning-tree cost512