Directed Graphs and Rectangular Layouts (Extended abstract)

Directed Graphs and Rectangular Layouts

(Extended abstract)

Adam L.Buchsbaum,Emden R.Gansner,and Suresh Venkatasubramanian

AT&T Labs–Research

Florham Park,NJ07932

{alb,erg,suresh}@https://www.wendangku.net/doc/3414466719.html,

Abstract.This paper deals with the problem,arising in practice,of drawing a di-

rected graph as a collection of disjoint,isothetic rectangles,where the rectangles

of the nodes of each edge must touch and where the placement of the rectangles

respects the ordering of the edges.It provides characterizations for those graphs

having the special type of rectangular layout known as a rectangular dual.It then

characterizes the st-graphs having rectangular layouts in terms of the existence of

certain planar embeddings and the non-existence of a particular subgraph.

1Introduction

We consider the problem of drawing a directed graph using disjoint,isothetic rectangles for the nodes,such that(u,v)is an edge in the graph if and only if the corresponding rectangles R u and R v touch,with R u above or to the left of R v.Figure1provides an example,showing a directed tree drawn both using the traditional,straight-line drawing and as a collection of rectangles.

The problem arose in the context of providing a graphical user interface to a re-lational database support system which allows users to model and administer their https://www.wendangku.net/doc/3414466719.html,ers can specify their own schemas using entity-relationship models[17].It is assumed that there are no cycles among the relationships in the schema.The system then displays the database entities as rectangular buttons.Clicking on a button provides information related to the corresponding entity type,such as detailed descriptions of the entity attributes or information about speci?c records.Experience indicates the bene?t of juxtaposing buttons that correspond to related entities.In addition,if the relationship is one-to-many,the button corresponding to the“many”entity is positioned below or to the right of the“one”entity,to emphasize this sense of directionality.There are no additional constraints.In particular,even if two entities are not related,their rectan-gles may abut.Viewing the database schema as the obvious directed tree,with entities as vertices and relations as edges,and allowing undirected cycles,leads to the graph drawing problem stated above.1

This generalizes the concept of rectangular layouts,in which an undirected graph G is drawn using disjoint rectangles for nodes,and two nodes are adjacent in G if and only if the two corresponding rectangles are adjacent.Without directionality,this[10,11,

RPremises RElement Premise Element Detail Port Gateway RCard IOC Trunk RPPort LPort RLPort Pvc

Fig.1.Traditional drawing and rectangular layout of tree.

3,12,6,9,14,7,15,16,5]and related problems,such as proximity drawing [2,8]and rectangle of in?uence drawings [13,4],have been extensively studied.In particular,there are effective characterizations for graphs having rectangular layouts,and some provable bounds on the size of the layouts.

In this paper,we explore what effect the directionality constraint has.In Section 2,we formally de?ne the problem and note some basic results.In Section 3,we consider the special case of rectangular duals,where the rectangles in a rectangular layout form one large rectangle.We give several characterizations for graphs having a rectangular dual.Then,in Section 4,we show that,for st-graphs,we have the same characterization as in the undirected case after accounting for one forbidden subgraph.

2Preliminaries

Throughout,we assume that all graphs are directed and connected,with no loops or multiple edges,and have at least 4vertices.

A (strong)rectangular layout of a graph G =(V,E )is a set R of isothetic rectan-gles whose interiors are pairwise disjoint,with an isomorphism R :V →R such that for any two vertices u,v ∈V ,the boundaries of R u =R (u )and R v =R (v )overlap non-trivially with

R (u )above or to the left of R (v )(1)

if and only if (u,v )∈E .Figure 2gives two graphs and sample rectangular layouts.

It is immediate that only acyclic graphs can have a rectangular layout.Thus,we assume that any graph is a directed acyclic graph (dag ).

A weak rectangular layout of G is a rectangular layout in which two nodes might touch even if there is no edge between the corresponding nodes.Note that,in Figure 1,layout (b)is a weak layout for both graphs (a)and (b)but a strong layout only for graph (b).The reader may have also noticed that the rectangular layout in Figure 1is weak,with the rectangles for nodes Detail and IOC touching but with no edge connecting them.

For purposes of construction,however,the distinction between strong and weak layouts does not matter,as shown by the following lemma.

c b

d a d a (a)a (b)d

b

c b c

Fig.2.Two graphs and associated rectangular layouts.The shaded region depicts a gap.Lemma 1.Graph G has a weak rectangular layout R if and only if G has some strong rectangular layout L .

Proof.The shearing argument given in Buchsbaum et al.[5]extends to the directed case.

Furthermore,the assumption that no two rectangles meet trivially at a corner can be relaxed.If rectangles R u and R v meet at a corner,we can perturb the boundaries by some small amount to make the boundary overlap non-trivial.The layout becomes weak if it was not already.Lemma 1shows that it can be made strong with only non-trivial boundary overlaps.

3Directed rectangular duals

For the case of rectangular layouts of undirected graphs,it has been useful (e.g.,see

[5,6,10–12])to consider the case where the rectangles form a partition of an enclos-ing rectangle.We take a similar approach here.A rectangular dual of a graph G is a rectangular layout in which the union of the rectangles is a rectangle.

Let G be a planar st-graph 2with embedding E .Let b r =(v 0=s,v 2,...,v n =t )be the right boundary path of E from s to t.Let b l =(u 0=s,u 2,...,u m =t )be the left boundary path of E from s to t.Any vertex not in b r or b l is an interior vertex.Add four vertices N,E,S,and W so that N is above and to the right of G ,W is above and to the left of G ,S is below and to the left of G ,and E is below and to the right of G .We say the boundary of E is embeddable if we can ?nd 0≤i ≤n and 0≤j ≤m such that G extended with the nodes N,E,S ,and W and the edges (N,v k ),0≤k ≤i ,(v k ,E ),i ≤k ≤n ,(W,u k ),0≤k ≤j ,and (u k ,S ),j ≤k ≤m ,is planar,and all of the vertices v k and u k have indegree and outdegree ≥2.

Theorem 1.A dag G has a directed rectangular dual if and only if G is a planar st-graph with an embedding E such that:

1.all interior faces are triangles

2.all interior vertices have indegree and outdegree ≥2.

3.the boundary of E is embeddable.

Proof.For necessity,the?rst two conditions are obvious.As for the third,if G has a rectangular dual,attaching4rectangles to the sides of the layout shows that the bound-ary is embeddable.

Let G be G extended with the vertices N,E,S,W,the edges from condition3), and the edges(N,E)and(W,S).Note that for any vertex v in G,its rightmost inedge and outedge form two edges of a triangle in G .The similar property holds for the leftmost inedge and outedge.

For suf?ciency,we show how to construct a regular edge labeling(REL)in the sense of He[6]on G .Given an REL,He’s construction produces a rectangular dual for the underlying undirected graph of G .It is simple to verify that the construction honors edge directions,thereby giving us a(directed)rectangular dual for G .

We now show how to construct the necessary REL.Recall that an REL is a partition of the directed edges3into2sets T1and T2such that,for each vertex v,the edges adjacent to v consist of the inedges of T1,followed by the inedges of T2,followed by the outedges of T1,followed by the outedges of T2,in the counterclockwise direction. Each of the four subsets must be non-empty.

We de?ne T1to include the leftmost outedge and the rightmost inedge of every vertex.Let T2be the complement of T1.To show that the four subsets of edges around a vertex are non-empty,it suf?ces to show that the leftmost inedge and the rightmost outedge of each vertex will be in T2.Let e=(v,w)be the rightmost outedge of v. Since v has outdegree at least2,e cannot be the leftmost outedge.In addition,if(u,v) is the rightmost inedge of v,we must have a face{u,v,w}with an edge(u,w),so e cannot be the rightmost inedge of w.Thus,e must be in T2.The symmetric argument holds for the leftmost inedge of v.

To see that the edges are appropriately ordered,let e=(v,w)be an outedge of v in T1.If e is not the leftmost outedge,let e =(v,u)be the next outedge to the left.Since e is in T1and e is not the leftmost outedge of v,we must have that e is the rightmost inedge of w.Since w has indegree at least2,there must be another inedge of w just to the left of e.This means that{v,u,w}forms a face and e is the rightmost inedge of u, so that e is in T1.This implies that any outedge to the left of an outedge in T1is also in T1,and therefor all T1are to the left of all T2outedges.A symmetric argument holds for the inedges.

Figure3(a)exhibits an embedded planar st-graph.Figure3(b)shows the edges be-longing to the sets T1and T2as de?ned in the theorem.

Once the four vertices N,E,S,W are attached,the amount of variation in con-structing an REL is limited by the following theorem.We say an edge is a middle edge if has the form shown by the dotted edge in Figure4.

Theorem2.If G is a dag satisfying the three conditions of Theorem1,then any edge

e satis?es exactly one o

f the following:

1.e is the leftmost inedge and/or the rightmost outedge of a vertex

Fig.3.(a)st-graph G(b)extended st-graph G with edge sets T1(dotted)and T2

2.e is the leftmost outedge and/or the rightmost inedge of a vertex

3.e is a middle edge

Proof.We have already observed,as part of the proof of Theorem1,that the?rst two sets are disjoint.To complete the proof,we simply have to show that any edge not in one of the?rst two sets must be a middle edge.This is almost immediate.If e=(u,v) is not the rightmost outedge of u nor the rightmost inedge of v,the face to the right of e must have be composed of the three edges(u,v),(u,w),(w,v)for some vertex w. Since the same must be true for face to the left of e,we see that e is a middle edge.

Theorem2shows that the only choices in constructing an REL,once the graph is embedded,concern where to put the middle edges.It is a simple observation that any middle edge can be in either T1or T2,so the number of distinct directed rectangular duals is2m,where m is the number of middle edges.

There is a simple suf?cient condition for property3)in Theorem1,namely,that,on the right and left boundaries of E,no vertex with indegree1appears after any vertex with outdegree1.Consider the right boundary b r=(v0=s,v2,...,v n=t)of E.Let k be the largest index so that v k has indegree<2.Let l be the smallest index so that v l has outdegree<2.Let i be any value k≤i≤l,and add the edges(N,v k),0≤k≤i and(v k,E),i≤k≤n.By hypothesis,the extended graph is planar.Each v k in G has indegree and outdegree at least1.If it has indegree1,the edge(N,v k)was added,so it

Fig.4.Middle edge

has indegree2in the extended graph.The analogous results hold for the outdegrees in b r,and for the left boundary of E.

This condition characterizes the biconnected graphs with directed rectangular du-als.(We consider a dag biconnected or not according to that property holding in the underlying undirected graph.)

Theorem3.A biconnected dag G has a directed rectangular dual if and only if G is a planar st-graph with an embedding E such that:

1.all interior faces are triangles

2.all interior vertices have indegree and outdegree≥2.

3.on the right and left boundaries of E,no vertex with indegree1appears after any

vertices with outdegree1.

Proof.We have already noted how the three properties,along with Theorem1,imply the existence of a directed rectangular dual.Conversely,if G has a directed rectangular dual,any vertex v on b r with indegree1must correspond to a rectangle on the top boundary.Any vertex u on b r appearing before v is therefore also on the top boundary with a rectangle to its right.Since G is biconnected,u cannot also lie along the bottom boundary,as then it would be an articulation point.Thus,u has a rectangle below it. This shows it has outdegree≥2.

It is simple to apply this result to the cases where the condition3)is trivially satis-?ed.

Corollary1.A biconnected dag G has a directed rectangular dual with a single rect-angle on both the right and left sides if and only if it satis?es properties1)and2)of Theorem3and all vertices of one boundary path have outdegree≥2,and all vertices of the other boundary path have indegree≥2.

Analogous results hold when the degree constraints are only applied to one side.

Theorem3clearly fails for non-biconnected graphs.A chain has a rectangular dual but condition3)does not hold.The next theorem,however,allows us to apply the the-orem to the block decomposition of non-biconnected graph.

Theorem4.A dag G has a rectangular dual if and only if

1.its block tree is a chain

2.each interior block has a directed rectangular dual with a single rectangle on both

the left and right sides

3.both the?rst and last blocks have rectangular duals,with a single rectangle on the

right and left side,respectively.

Proof.Given the3conditions,joining the rectangular duals in the order speci?ed by the block tree chain,abutting or identifying the side rectangles as necessary,gives a dual for the whole graph.On the other hand,if G has a rectangular dual,any rectangle R v corresponding to a cut point v must be as wide or as high as the whole rectangle. By a90?rotation,we can assume it is as high.By dividing the rectangular dual at R v, splitting R v into2rectangles if both its indegree and outdegree are greater than1,we end up with rectangular duals for2proper subgraphs of G,and can complete the proof by induction using Theorem3for the base case.

4st-Graphs with rectangular layouts

As with the undirected case,any graph with a(directed)rectangular layout is a subgraph of some graph with a(directed)rectangular dual.Thus,the results from Section3can help us characterize certain classes of graphs with rectangular layouts.

We start with some de?nitions and lemmas.We let F denote the graph shown in Figure5(a).From the next lemma,we also refer to F as forbidden.

Lemma2.Let G be a biconnected planar st-graph with an embedding such that all interior faces are triangles.The condition that all interior vertices have indegree and outdegree≥2is equivalent to G not containing a subgraph isomorphic to F.

(a)(b)

(c)

Fig.5.(a)The forbidden graph F.(b)Quadrilateral face(c)Triangulation of a quadrilateral face Proof.If G has F as a subgraph,we can assume,by symmetry,that its embedding is as shown in Figure5(a).Consider the set formed by u and all vertices within the quadrilateral,and pick one w with the minimum y coordinate.Then the only possible outedge is(w,v),and w has outdegree1.

To show the opposite direction,we can assume,without loss of generality,that G has a vertex u with outdegree1.Let(u,v)be the unique outedge.The two triangular faces common to(u,v)must have the form(w,u),(w,v),(u,v)and(z,u),(z,v),(u,v). Therefore,G has the subgraph(w,u),(w,v),(z,u),(z,v)isomorphic to F.

For a planar embedding of a graph,a?lled triangle is de?ned to be a length-3cycle with at least one vertex inside the induced region.

Lemma3.Let E be an embedding of a planar st-graph G.If E has a?lled triangle,G has an interior vertex with outdegree or indegree1.

Proof.Consider a?lled triangle with edges(v,u),(v,w)and(u,w).If there is any vertex inside the triangle whose y coordinate is the same or less than that of u,pick the one x with the smallest y coordinate.Then the only possible outedge of x is(x,w)and x has outdegree1.A symmetric argument holds if any vertex inside the triangle has y coordinate the same or more than that of u.

We can now characterize planar st-graphs having a rectangular layout.

Theorem5.Let G be a planar st-graph.G has a rectangular layout if and only if G has no subgraph isomorphic to F and has an embedding with no?lled triangles. Proof.F has no rectangular layout,so it cannot be a subgraph of any graph having one.In addition,if G has a rectangular layout,it is a subgraph of a graph H with a rectangular dual.By Theorem1,H has an embedding with all interior faces triangles and all interior vertices having outdegree and indegree greater than1.By Lemma3,the embedding for H,and hence G,has no?lled triangle.

To show the converse,?rst assume that G is biconnected and let E be an embedding of G.For each non-triangular face F,let n be the length of the longest boundary path (i.e.,the path has n+1vertices).If n=2,then F must look like Figure5(b).Add a single vertex and four edges to form the graph in Figure5(c).The added edges are dotted.To maintain a valid planar embedding of an st-graph,we need to vertically shift down node c and all nodes reachable from it by a directed path by a uniform amount, as indicated in the?gure.

If n>2,insert n?2vertices to form a path in the interior from the top vertex of F to the bottom.Alternately connect the vertices along the longest boundary path to the inserted vertices as indicated in Figure6(a).The added edges are dotted.Similarly triangulate the shorter side,connecting the highest boundary path node to all of the remaining added vertices.None of these operations can introduce a?lled triangle,a cut vertex,or a subgraph isomorphic to F.

After processing all non-triangular faces,the resulting graph is still biconnected and acyclic,with a planar embedding with all interior faces triangular.In addition,there is no subgraph isomorphic to F.Thus,by Lemma2,every interior vertex has indegree and outdegree at least2.

To be able to invoke Theorem3,we need to make sure the graph satis?es property 3.For a given boundary path,if there is any node of indegree1appearing after a node of outdegree1,let n be the number of vertices on the boundary path from s to t.The cases where n<4are trivial,so we can assume n≥4.Let the vertices on the boundary path be s,v,...,w,t.Add n?3new vertices on a path from v to w.Then triangulate the face as indicated in Figure6(b)for the case n=6.As usual,the added edges are dotted.This process eliminates any vertices of outdegree1before a vertex of indegree 1,and maintains the hypotheses of the theorem.

We now can apply Theorem3to get a rectangular dual for the extended graph. Removing the rectangles corresponding to added vertices gives a rectangular layout for G.

Fig.6.(a)Triangulation of a non-quadrilateral face(b)Triangulating a boundary path

Finally,if G is not biconnected,we can use induction,with the base case of a bi-connected graph dealt with above.Pick a cut vertex v in G.Possibly replicating v if both its indegree and outdegree are greater than1,we can split G at v into two smaller st-graphs,each satisfying the hypotheses of the theorem.By induction,we get rectan-gular layouts for each.Necessarily,in any rectangular layout of an st-graph,the source and sink nodes correspond to the upper left and lower right rectangles,respectively.As in Theorem4,we can join the layouts together,identifying rectangles if necessary for replicated nodes,to obtain a rectangular layout for G.

Except for the forbidden subgraph,Theorem5is identical to Corollary3.5in[5]. Note that if all face boundary paths have length at least2,there can be no?lled triangles. The example of Figure7(a)shows how the triangulation used in the proof of the theorem fails in the case of?lled triangles.The graph is the same that of Figure5(c)but we assume it also has an edge(a,d),giving the?lled triangle(a,d),(a,c),(c,d).When the auxiliary node(e)and(dotted)edges are added,we have a forbidden subgraph (a,d),(e,d),(a,c),(e,c).

Both conditions of the theorem are necessary in order to have a rectangular layout. Figure7(b)shows an example of an st-graph with no forbidden subgraph all of whose embeddings have a?lled triangle.Figure7(c)shows an example of an st-graph with a forbidden subgraph but no?lled triangle.Of course,neither has a rectangular layout.

We believe Theorem5can be extended to general dags as follows:

Conjecture1.Let G be a dag.G has a directed rectangular layout if and only if G has no subgraph isomorphic to F and has an upward planar embedding with no?lled triangles.

Fig.7.(a)Filled triangle yielding F(b)All embeddings with?lled triangle(c)Subgraph F but no?lled triangle

The idea of the proof should be the same as that for Theorem5but the bookkeeping is more complex.

5Summary and future work

If this paper,we considered the extension of the problem of rectangular layouts to di-rected graphs.We presented characterizations for which graphs have rectangular duals, and which st-graphs admit rectangular layouts.

As for future work,we have already noted the conjecture at the end of the previous section.In analogy with the undirected case,there are many avenues open for further research.In particular,given the original motivation for the problem,it would be nice to have a layout algorithm with provably tractable running time and provable area or width,bounds similar to the results presented by Buchsbaum et al.[5]for the undirected problem.

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英语中很多由动词转化而成的名词、以及动名词、非谓语动词等，汉译时可将它们转换成动词。 ㈣.补词 是指原文已有某种含义但未用词汇直接表达，译文中需将这些含义补充进去，这样才更通顺易读，如：英语中数词与名词之间没有量词，而译成汉语时可酌情增加。 ㈤.省略 是指原文中某些词在译文中省略不译，只要并不影响意义的完整。如：上面讲的汉语“量词”，译成英语时则可以省略；又如：英语中大量使用物主代词而汉语中往往省略不用。 ㈥.并列与重复

英语在表达重复含义的并列结构中常采用共享、替代、转换等形式来避免重复，而汉语却常常有意重复表达以加强文字的力度，如：英语的物主代词替代前面的名词，短语动词只重复介词而省略主动词，汉译时可考虑重复表达。 二、句子结构方面 子结构方面的翻译技巧，主要有三种类型：语序类、组合类和转换类。 ㈠.语序类 1. 顺译法与逆译法 英语时间状语可前可后。不仅如此，英语在表达结果、条件、说明等定语从句、状语从句也很灵活，既可以先述也可以后述。而汉语

表达往往是按时间或逻辑的顺序进行的，因此，顺译法也罢逆译法也罢，其实都是为了与汉语的习惯相一致。英语表达与汉语一致的就顺译，相反的则逆译。 有时候顺译法与逆译法的差别,就象前面谈的正译与反译,依译者的爱好而定。 2. 前置法 英语中较短的限定性定语从句、表身份特征等的同位语在译成汉语时，往往可以提到先行词（中心词）的前面。 3. 分起总叙与总起分叙 长句子和句子嵌套现象在英语中比较普遍，这是因为英语的连词、关系代词、关系副词等虚词比较活跃、生成能力强，可构成并列句、复合句以及它们的组合形式。

翻译中词义的选择

Run的词义选择 1.run across 2.run away 3.run back over 4.run counter to 5.run down 6.run in 7.run into 8.run into debt 9.run off 10.run out 11.run over 12.run through 13.run up 14.run to seed 15.run a race 16.run for presidency 17.run to sb’s aid 18.run agroun 19.run wild 20.run to extremes 21.run the streets 22.run a car 23.run a factory 24.run message 25.run the water off 26.run oil 27.run on 28.run sb. into difficulties 29.run the risk of 30.run one’s eyes down a list 31.run a simile too far 32.run a fever 33.run a sword through sb. 34.run a thread through an eyelet 35.run one’s head into wall 36. The idea runs in his mind. 37.The street runs north. 38.The news runs rapidly in the town. 39.The machine runs well. 40.The river runs east. “上”的不同含义 1.上班 2.上学 3.上岗 4.上报 5.上场 6.上当 7.上冻 8.上访 9.上纲 10.上钩 11.上光 12.上轨道 13.上火 14.上进 15.上课 16.上口 17.上路 18.上门 19.上马 20.上年纪 21.上任 22.上台 23.上膛 24.上相 25.上刑 26.上演 27.上账 28.上阵 29.上瘾 30.上街 31.上朝 32.上书 33.干部要能上能下 34.随函复上1元钱邮票一张。 35.这一局你上。 36.一连上了好几道菜。 37.行李还没上架。 38.他正在给门上漆。 39.这事已经上了电视。 40.表该上了。 1

无线传感器网络中英文对照外文翻译文献

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无线传感器网络的应用研究

1武警部队监控平台架构介绍与设计 1.1监控系统的系统结构 基站监控系统的结构组成如上图所示，主要由三个大的部分构成，分别是监控中心、监控站点、监控单元。整个系统从资金、功能以及方便维护性出发，我们采用了干点加节点方式的监控方法。 监控中心（SC）：SC的定义是指整个系统的中心枢纽点，控制整个分监控站，主要的功能是起管理作用和数据处理作用。一般只在市级包括（地、州）设置相应的监控中心，位置一般在武警部队的交换中心机房内或者指挥中心大楼内。 区域监控中心（SS）：又称分点监控站，主要是分散在各个更低等级的区县，主要功能是监控自己所负责辖区的所有基站。对于固话网络，区域监控中心的管辖范围为一个县/区；移动通信网络由于其组网不同于固话本地网，则相对弱化了这一级。区域监控中心SS的机房内的设备配置与SC的差不多，但是不同的是功能不同以及SS的等级低于SC,SS的功能主要是维护设备和监控。 监控单元（SU）：是整个监控系统中等级最低的单元了，它的功能就是监控并且起供电，传输等等作用，主要由SM和其他供电设备由若干监控模块、辅助设备构成。SU侧集成有无线传感网络微设备，比如定位设备或者光感，温感设备等等。 监控模块（SM）：SM是监控单元的组成部分之一，主要作用监控信息的采集功能以及传输，提供相应的通信接口，完成相关信息的上传于接收。

2监控系统的分级管理结构及监控中心功能 基站监控系统的组网分级如果从管理上来看，主要采用两级结构：CSC集中监控中心和现场监控单元。CSC主要设置在运营商的枢纽大楼，主要功能为数据处理，管理远程监控单元，对告警信息进行分类统计，可实现告警查询和存储的功能。一般管理员可以在CSC实现中心调度的功能，并将告警信息进行分发。而FSU一般针对具体的某一个基站，具体作用于如何采集数据参数并进行传输。CSC集中监控中心的需要对FSU采集的数据参数进行报表统计和分析，自动生产图表并为我们的客户提供直观，方便的可视化操作，为维护工作提供依据，维护管理者可以根据大量的分析数据和报表进行快速反应，以最快的速度发现网络的故障点和优先处理点，将人力资源使用在刀刃上。监控中心CSC系统的功能中，还有维护管理类，具体描述如下： 1）实时报警功能 该系统的报警功能是指发现机房里的各种故障后，通过声音，短信，主界面显示的方式及时的上报给操作者。当机房内的动力环境，空调，烟感，人体红外等等发生变量后，这些数据通过基站监控终端上传到BTS再到BSC。最后由数据库进行分类整理后存储到SQLSEVRER2000中。下面介绍主要的几种报警方式： 2)声音报警 基站发生告警后，系统采集后，会用声卡对不一样的告警类别发出对应的语音提示。比如：声音的设置有几种，主要是以鸣叫的长短来区分的。为便于引起现场维护人员的重视紧急告警可设置为长鸣，不重要的告警故障设置为短鸣。这样一来可以用声音区分故障的等级，比方某地市的中心交换机房内相关告警声音设置，它的开关电源柜当平均电流达到40AH的时候，提示声音设置为长鸣，并立即发生短信告警工单。如果在夜晚机房无人值守的情况下：

无线传感器网络中文

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考研翻译中词义的选择和引申

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