对于具有500个网络节点和4000条链路的大型网络，请考虑以下关于链路状态和距离矢量路由协议的以下三个陈述。
[S1] The computational overhead in link state protocols
is higher than in distance vector protocols.
[S2] A distance vector protocol (with split horizon)
avoids persistent routing loops, but not a link
state protocol.
[S3] After a topology change, a link state protocol
will converge faster than a distance vector
protocol.关于S1，S2和S3，下列哪一项是正确的?
(A) S1，S2和S3都是正确的。
(B) S1，S2和S3均为假。
(C) S1和S2为真，但S3为假
(D) S1和S3为真，但S2为假答案： (D)
解释：
Link-state:
Every node collects complete graph structure
Each computes shortest paths from it
Each generates own routing table
Distance-vector
No one has copy of graph
Nodes construct their own tables iteratively
Each sends information about its table to neighbors 资料来源：http://www.cs.cmu.edu/~srini/15-441/S05/lectures/10-Routing.ppt
[S1] The computational overhead in link state protocols
is higher than in distance vector protocols.
[S2] A distance vector protocol (with split horizon)
avoids persistent routing loops, but not a link
state protocol.
[S3] After a topology change, a link state protocol
will converge faster than a distance vector
protocol.S1显然是正确的，因为在链接状态中，所有节点都为整个网络图计算最短路径。
S3也是正确的，因为距离矢量协议已计入无穷大问题，并且收敛速度较慢。
S2为假。在距离矢量协议中，带有毒药反向的水平分割减少了形成回路的机会，并使用最大跳数来解决“无穷计数”问题。这些措施可以避免在某些(但不是全部)情况下形成路由环路
这个问题的测验