2011数学建模B题
2011年数学建模论文 B 题
承诺书我们仔细阅读了中国大学生数学建模竞赛的竞赛规则.我们完全明白，在竞赛开始后参赛队员不能以任何方式（包括电话、电子邮件、网上咨询等）与队外的任何人（包括指导教师）研究、讨论与赛题有关的问题。
我们知道，抄袭别人的成果是违反竞赛规则的, 如果引用别人的成果或其他公开的资料（包括网上查到的资料），必须按照规定的参考文献的表述方式在正文引用处和参考文献中明确列出。
我们郑重承诺，严格遵守竞赛规则，以保证竞赛的公正、公平性。
如有违反竞赛规则的行为，我们将受到严肃处理。
我们参赛选择的题号是（从A/B/C/D中选择一项填写）： B我们的参赛报名号为（如果赛区设置报名号的话）：所属学校（请填写完整的全名）：四川文理学院参赛队员(打印并签名) ：1. 高陆2. 肖皓华3. 吕洋琴指导教师或指导教师组负责人(打印并签名)：李爱民日期： 2011 年 9 月 12 日赛区评阅编号（由赛区组委会评阅前进行编号）：编号专用页赛区评阅编号（由赛区组委会评阅前进行编号）：赛区评阅记录（可供赛区评阅时使用）：评阅人评分备注全国统一编号（由赛区组委会送交全国前编号）：全国评阅编号（由全国组委会评阅前进行编号）：关于交巡警服务平台设置与调度的优化模型摘要本文旨在研究交巡警服务平台的设置与调度问题，根据不同的问题分别建立了与之相应的数学模型。
问题一以中心城区A为主要研究范围，而问题二则主要研究全市（主城六区A，B，C，D，E，F）交巡警服务平台的设置与调度问题。
为解决问题一的三个子问题，我们在认真研究该城区交通网络及现有的20个交巡警服务平台设置情况的基础上，利用题中所给附件2中列出的关于城区A的92个节点的横纵坐标这一具体数据，得出了各个节点间的距离（程序实现见附录2）。
针对问题一的子问题一，即为城区A的各交巡警服务平台分配管辖范围，因要考虑到当在其管辖范围内出现突发事件时，该服务平台的交巡警能尽量在3分钟内到达事发地。
而在此3分钟内，警车所经过的路程不超过3km（警车时速为60km/h）。
2011高教社杯全国大学生数学建模竞赛B题参考答案
交巡警服务平台的设置与调度优化分析摘要本文以实现警察的刑事执法、治安管理、交通管理、服务群众四大职能为宗旨，利用有限的警务资源，根据城市的实际情况与需求合理地设置了交巡警服务平台、分配各平台的管辖范围及调度警务资源。
并分别对题目的各问，作了合理的解答。
问题一：（1）、根据题目所给数据，确定各节点之间的相邻关系和距离，利用Floyd 算法及matlab编程求出两点之间的最短距离，使其尽量满足能在3分钟内有交巡警平台警力到达案发结点的原则，节点去选择平台，把节点分配给离节点距离最近的平台管辖，据此，我们得到了平台的管辖区域划分。
（2）、我们对进出该区的13条交通要道实现快速全封锁的问题，我们认定在所有调度方案中，某种方案中耗时最长的的围堵时间最短即最佳方案，利用0-1变量确定平台的去向，并利用线性规划知识来求解指派问题，求得了最优的调度方案。
（3）、在确定增添平台的个数和具体位置的问题中，我们将尽量保证每个节点都有一个平台可以在三分钟内到达作为主要原则来求解。
我们先找出到达每个平台的时间都超过三分钟的节点，并尝试在这些节点中选取若干个作为新的平台，求出合理的添加方案。
问题二：（1）、按照设置交巡警服务平台的原则和任务，分析现有的服务平台的设置是否合理，我们以各区覆盖率作为服务平台分布合不合理的评价标准，得到C、D、E、F区域平台设置不合理。
并尝试一些新的设置方案使得设置更为合理，最后以覆盖率最低的E区为例，使用一种修改方案得到一个比原方案更合理的交巡警服务平台的设置方案。
（2）、追捕问题要求在最快的时间内抓到围堵罪犯，在罪犯和警察的行动速度一致的前提假设下，我们先设定一个具体较小的时间，编写程序检验在这个时间内是否可以成功抓捕罪犯，不行则以微小时间间隔增加时间，当第一次成功围堵时，这个时间即为最佳围堵方案。
关健字： MATLAB软件，0-1规划，最短路，Floyd算法，指派问题一、问题重述“有困难找警察”，是家喻户晓的一句流行语。
2011全国数学建模B题 交巡警服务平台的设置与调度
2011高教社杯全国大学生数学建模竞赛题目（请先阅读“全国大学生数学建模竞赛论文格式规范”）B题交巡警服务平台的设置与调度“有困难找警察”，是家喻户晓的一句流行语。
警察肩负着刑事执法、治安管理、交通管理、服务群众四大职能。
为了更有效地贯彻实施这些职能，需要在市区的一些交通要道和重要部位设置交巡警服务平台。
每个交巡警服务平台的职能和警力配备基本相同。
由于警务资源是有限的，如何根据城市的实际情况与需求合理地设置交巡警服务平台、分配各平台的管辖范围、调度警务资源是警务部门面临的一个实际课题。
试就某市设置交巡警服务平台的相关情况，建立数学模型分析研究下面的问题：（1）附件1中的附图1给出了该市中心城区A的交通网络和现有的20个交巡警服务平台的设置情况示意图，相关的数据信息见附件2。
请为各交巡警服务平台分配管辖范围，使其在所管辖的范围内出现突发事件时，尽量能在3分钟内有交巡警（警车的时速为60km/h）到达事发地。
对于重大突发事件，需要调度全区20个交巡警服务平台的警力资源，对进出该区的13条交通要道实现快速全封锁。
实际中一个平台的警力最多封锁一个路口，请给出该区交巡警服务平台警力合理的调度方案。
根据现有交巡警服务平台的工作量不均衡和有些地方出警时间过长的实际情况，拟在该区内再增加2至5个平台，请确定需要增加平台的具体个数和位置。
（2）针对全市（主城六区A，B，C，D，E，F）的具体情况，按照设置交巡警服务平台的原则和任务，分析研究该市现有交巡警服务平台设置方案（参见附件）的合理性。
如果有明显不合理，请给出解决方案。
如果该市地点P（第32个节点）处发生了重大刑事案件，在案发3分钟后接到报警，犯罪嫌疑人已驾车逃跑。
为了快速搜捕嫌疑犯，请给出调度全市交巡警服务平台警力资源的最佳围堵方案。
附件1：A区和全市六区交通网络与平台设置的示意图。
附件2：全市六区交通网络与平台设置的相关数据表（共5个工作表）。
交巡警的服务平台的设置与调度摘要正在整理……一、问题重述……二、问题分析……三、模型的假设^四、符号说明^五、模型的建立与求解问题一：（1）各交巡警服务平台的管辖范围，尽量在分钟内到达事发地，实质上是求最短路径问题。
2011数学建模(美赛)b题
Minimizing the Number of repeatersIntroductionVery high frequency (VHF) is the radio spectrum，whose frequency band ranges from 30MHz to 300MHz. VHF is always used for radio stations and television broadcasts. In addition, it is also used by signal transmission of sea navigation and aviation. Because the radio spectrum of VHF is transmitted through straight lines, a signal is easily influenced by geographical factors easily. Thus, signals become weak when it is transmitted and some low-power users need repeaters to amplify them and increase the transmission distance. We consider the situation in which every two repeaters are too close or the separate frequency is not far enough apart which can interference with each other. In order to mitigate the interference caused by the nearby repeaters, this paper employs a continuous tone-coded squelch system (CTCSS). We associate to each repeater a separate subaudible tone，that is, the subaudible tone (67Hz-250.3Hz) is added to VHF. In this way, repeaters recognize signals attached to the same subaudible tones just like secret keys. In this system, the nearby repeaters can share the same frequency pair. When users send the signals at one frequency, different repeaters with subaudible tones can recognize signals from the users the same subaudible tone. If the users in a certain area contact with each other, we should consider the signal’ s coverage area of the users and the repeaters. As long as the users’ signals are accepted by repeaters, the signals could be amplified to transmit farther. At the same time, the repeaters attached with the subaudible tones could only recognize the users with the same subaudible tones. Hence, we can consider repeaters corresponding to the number of the users, which leads to the problem of frequency channel. When the number of users in this area increases, we can add repeaters. If two repeaters have different subaudible tones, they would not communicate with each other. Thus, we should consider the problem of how the repeaters communicate with each other when they have different subaudible tones. In the mobile communication system，the spectrum is influenced by many factors such as reflex，diffraction and dispersion. Therefore, when the radio spectrum transmits in the mountainous area，we should still consider the factors above.Repeaters[4]Repeaters are a type of equipment which can amplify signals，make up the deamplification signals and support far distance communication.CTCSS[5]CTCSS(Continuous Tone Controlled Squelch System ) is short for subaudible tones, whose frequency ranges from 67Hz to 250.3Hz. It is added to the radio spectrum to make the signal carry with a unique secret key.AssumptionThe users in the area is uniform distributedThe signal of the radio spectrum in the area can’t be effected by environmentIn a certain period of time there are a small number of users removingAll repeaters have the same standardAnalysis and solution of the model to the first problemThe problem is to find a least number of repeaters in an area of radius 40 miles so that the users in this area can communicate with each other. Considering that the given area is flat, we assume that the signal ofeach repeater covers a circular area and the repeater lies in the center of the circle. The following Figure 1 shows the relationship of three adjacent repeaters.CFor case B of Figure 1, if three circles are tangent to each other, then we find that the center area cannot be covered by the singles. In order to make the signal cover the triangle area, we have to consider adding a For case C, if the intersection of three circles is not null, similar to case B, we also have to add another repeater. Thus, it is easy to find that case A, comparing with cases B and C, is optimal. Thus, we obtain the largest covering area When linked hexagons, as shown in Figure 2. Obviously, it looks like a honeycomb structure. In fact, the honeycomb pattern is one of the most efficient arrangement for radio spectrum. It transmits by the wireless medium of microwave, satellites and radiation. The structure has a feature of point-to-point transmission or multicast. It is widely used in UN Urban Network, Campus Network and Enterprise Network.Figure 2. some circles intersecting together form the closely linked hexagons Now we have a circle with radius of 40 miles. Then we analyze the distances of signals from users and repeaters covering in the circle. Because the differences for the users and repeaters in energy and height, they have different covering distances. We calculate the distances with the theory of space loss. The formula[6]is1288.120lg 20lg 40lg LM F h h d =+-+,LM the wireless lossF the communication working frequency(MHz)1h the height of the repeater (m)2h the height of the user(m)d the distance between the user and repeater(km) We assume that 150F MHZ =,1 1.5h m = and 230h m =, under the condition of the cable loss and antenna gain, we obtain the system gain()(1,21,2)i j SG Pt PA RA CL RR i j =+-++==.The system gain is the allowed decay maximum of the signal from the users to repeaters. If the system gain value is higher than the wireless loss, the users could communicate with each other. Reversely, the users could not communicate. We make the system gain value equals to the wireless loss, thus, we get the extremity distance between the user and repeater. Then we haveSG LM =We choose a typical repeater and the user facility. Thus, the parameters [6] and data of the repeaters are as followsThe transmitting power 120(43)Pt W dBm =The receiving sensitivity 1116RR dBm =-The antenna gain of the repeaters 9.8RA dB =The cable loss 2CL dB =The parameters of the interphoneThe transmitting power 24(36)Pt W dBm =The receiving sensitivity 2116RR dBm =-The antenna gain of the interphone 0PA dB =The system gain of the system from users to repeaters 1144.2SG dB =. Thus, we get the sending distance from the users to repeaters 113.8d km =. Prove in the same way, we have the system gain of the system from the repeaters to users 2151.2SG dB =, the sending distance from the repeaters to the users 220.7d km =According to the sending distance 113.8D km = between user and repeater as well as the property of regular hexagon, we calculate the distance between two repeaters. We obtain that 223.09D km =, which is described in Figure 3. Because 2D is shorter than 2'D , users in this area cannot communicate with each other. Thus, we consider the sending distance 2'D between two repeaters firstly. Then we calculate the distance between the user and the repeater again shown in Figure 4. Finally, we get that 1'12.4D km =.Figure 3. the calculation distance according to the sending distance from users to repeaters.Figure 4. the calculation distance according to the sending distance from repeaters to the users.According to the calculated distance 12'12.4'21.45D km D km ==, we know that the given circle has a radius of 40 miles. We firstly consider the signals ’ covered area of the repeaters. Thus, we get the distribution of the repeater stations in this area showed in Figure 5. The number of repeater stations is 37. However, we need to decide the amount of repeaters distributing in one station.channel (the signaling channel between two points to transmit and receive signals) to transmit signals. Hence, we need 27 frequency channels [2] to maintain the normal communication.In order to avoid the interference about the close frequency between two repeaters, we arrange each repeater 10 frequency channels. We have121145.0145.03145.06145.09145.6145.63145.66145.69146.2146.23146.26146.29146.8146.83146.86146.89147.4147.43147.46147.49Mhz Mhz MHz MHz Mhz Mhz MHz MHz pl r Mhz Mhz MHz MHz Mhz Mhz MHz MHz Mhz Mhz MHz MHz r ⎧⎧⎪⎪⎪⎪⎪⎪⎨⎨⎪⎪⎪⎪⎩⎩（）233145.12145.15145.72145.75()()146.32146.35146.92146.95147.52147.55MHz MHzMHz MHz pl r pl MHz MHz MHz MHz MHz MHz ⎧⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩ Here, n is the number of repeaters.In this method of distribution ，we ensure that the signal could still be recognized after transmission. We associate to each repeater a subaudible tone and the users need to use the same tone to receive the corresponding signal. We suppose each repeater station have the same repeaters attached with different subaudible tones. In this way, we guarantee the signals transmitting in this zone without interference. Because when one user sends a signal with a specific frequency, the repeater could send the signal after adding or subtracting 600 KHz. However, our frequency channels cover the whole scope of the frequency. Thus, the signal can be transmitted in this zone.Finally, we calculate the number of the repeaters in a repeater station and obtain the number is 3. Thus, the total number of the repeaters is 3*37111=.When the number of users in this zone increases to 10000, we consider the problem as the first model. In this situation, each repeater station should cover 10000/37270.3= users. Hence, we need 270 frequency channels to maintain the normal communication. Since the number of the channels is too large, it is wasteful to use 10 frequency channels for the first problem. Thus, we consider assigning each repeater station 30 channels. Furthermore, we get 9 repeaters. However, for the frequency rand ranging from145MHz to 148MHz, the channel changes to 11.1KHz, which leads to the channels interfering with each other. Hence, we make use of the CTCSS system to distribute the 9 repeaters different PL tones. We can build the repeaters which can transmit the same frequency and have different tones.11145145.03145.06145.09145.012145.015145.6145.63145.66145.69145.72145.75()146.2146.23146.26146.29146.32146.35146.8146.83146.86146.89146.92146.95147.4147.43147.46147.49147.52147.55r mhz pl ⎧⎪⎪⎪⎨⎪⎪⎪⎩1'1'145145.03145.06145.09145.012145.015145.6145.63145.66145.69145.72145.75()146.2146.23146.26146.29146.32146.35146.8146.83146.86146.89146.92146.95147.4147.43147.46147.49147.52147.55r mhz pl ⎧⎪⎪⎪⎨⎪⎪⎪⎩Thus, we calculate the number of the repeaters in a repeater station is 270/309=. Then the total number of the repeaters is 9*37333=.The model of the line-of-sight propagation considering the effect ofthe mountainsWe search some information on how to build the repeaters at the top of the mountains. According to the factors influencing the positions of the repeaters, we establish a model to simulate these impact factors of transmission of VHF radio spectrum.When repeaters are installed at the tops of the mountainous, the positions of the repeaters are related to the height of the antenna, its coverage radius, the repeater power and antenna gain. Thus, it is difficult to build the communication network. In order to build communication network well, we should do lots of experiments to ensure the positions of the repeaters according to actual geomorphic environment.Since mountains have different heights, we mainly consider three cases. Case 1 is that the heights of the mountains are 15m below, case 2 requires that the heights ranges from 15 to 30m and the last one is 30m above.The Egli modelThis model considers the height of the mountains below 15m. We assume that the mountains in this zone have no larger peaks, that is, this zone is a medium rolling terrain.This model is based on the data of the mobile communication, which is established by Federal Communications Commission (FCC). It is an empirical equation which is summarized from the data of the irregular terrain. This model based on the barrier height is applied to the VHF radio spectrum and the irregular terrain. It demands the barrier height above 15m. When the barrier height is under 15m ，we modify the model to verify the modified factor T C . The loss of the spectrum [1] equation is218820lg 40lg 20lg 20lg T LM F d h h C =++---.Here, we assume that d is the distance between the two antennas (m), h ∆is the height of thetopography. If we use b h to denote the practical height of the sending signal antenna, o h to denote the least effective height of the antenna and m h the practical height of the receiving signal antenna, then theeffective height of the sending signal antenna 1h satisfies1()2b o h h h m +=, and the effective height of the receiving signal antenna 2h satisfies2()2m o h h h m +=, 100-10-20-301020305070100200300500t h e m o d i f y i n g f a c t o r s K /d B /h mFigure 6[1]. the range of the modifying factor. We obtain the relationship between the height of the topography and the modifying factor from the empirical data. Furthermore, we get the equation with respect to h ∆and T C .C 1.6670.1094h25150T MHz F MHz =-∆<< C 2.250.1476h150162T MHz F MHz =-∆<< C 3.750.2461h 450470T MHz F MHz =-∆<<This model for irregular area is fit for the frequency ranging from 40 to 450MHz. When the frequency is higher than 25MHz or lower than 400MHz and the distance between two antennas is less than 64km, the error would be very small. Through the model we can evaluate the value of the wireless loss and the number of the repeaters.Figure 7 describes the positions of the mobile station, repeater and the barrier. Next, we introduce the concept of the clearance.Figure 7．The schematic of the clearanceT the position of the mobile stationR the position of the repeater1d the distance between the mobile station and the barrier2d the distance between the repeaters and the barrierAssume that the line HD is perpendicular to line RT, which is called clearance showed in Figure 7. Because the distance between the two antennas is very far, thus, the HD is short. Then we can substitute the hd for HC . If the radius of the first Fresnel region (the region is used to evaluate the transmission energy of the video spectrum.) is 1F , we regard 1/HC F as the relative clearance.The equation [2] of the radius of the first Fresnel region is12112d d F d d λ=+where λ is a parameter.When the radio spectrum transmits ，there are always many barriers such as constructions, trees and peaks blocking the spectrum. If the height of the barrier has not reached the first Fresnel zone ，the barrier would have little influence to the receiving frequency level. However, when it is in the zone, it will cause the added losses (the power losses of the sending power relative to the receiving power) to decrease the receiving electrical level. The diffraction losses /dB T h e d i f f r a c t i o n l o s s e s /d BFigure 7. The relationship between diffraction losses and clearance [1].The relationship between the added losses and the clearance caused by the barriers is showed in Figure 7. When the height of the barrier is under the line RT and the relative clearance is larger than 0.5，the added losses changes around 0db. In this situation，the practical receiving electrical level approaches the value of the space loss. We can get the value of the clearance HC is less than0.557F or a negative value. It may1hinder the transmission of direct wave. Thus, we should make the barriers lie below the line RT. Strengths●In the first model, we distribute each repeater 5 frequency channels, meanwhile the different repeatershave different PL tones. Thus, under the condition of avoiding the interference of repeaters with each other, we control the number of frequency channels least to make the transmission more efficient.●The model is established when the users are uniformly distributed. When the number of users increases,the number of repeaters increases. Thus, this model applies the zone where the users are unevenlydistributed.●The Egli model is a model considering the modifying factors, which make the mountains areas problembe easily understood.Weaknesses●In the signal’s coverage area of the repeaters, we assume that each channel only has one user. However,in the practical situation, there may not be one user. That is to say, we have wasted the channel.●Our model belongs to fixed channels distribution strategies, the larger number of the users, the largernumber of the channels. It leads to channel interference with each other when channel bandwidth is less than 8.3MHz. Thus, our model only suits for less number of users.●Considering the mountains environment is complex, in our model, we only consider one mountaineffecting the transmission of radio spectrum.References[1] Yao Dongping, Huang Qing and Zhao Hongli, Digital Microwave Communication, Beijing: Beijing Jiaotong University Press, 2004.7.[2] Theodore S. Rappaport, Wireless Communications: Principles and Practice, Second Edition, Prentice Hall PTR，2006.7[3] DeWitt H.Scott, Michael Krigline, Successful Writing for the Real World, Foreign Language Teaching and Research Press, 2009.2[4] /wiki/Repeater, 2011.2.12[5] /wiki/CTCSS, 2011.2.12[6] /view/2074265.htm,2012.2.14。
2011年数学建模竞赛B题参考答案(只做了一半)
2011高教社杯全国大学生数学建模竞赛题目（请先阅读“全国大学生数学建模竞赛论文格式规范”）A题城市表层土壤重金属污染分析随着城市经济的快速发展和城市人口的不断增加，人类活动对城市环境质量的影响日显突出。
对城市土壤地质环境异常的查证，以及如何应用查证获得的海量数据资料开展城市环境质量评价，研究人类活动影响下城市地质环境的演变模式，日益成为人们关注的焦点。
按照功能划分，城区一般可分为生活区、工业区、山区、主干道路区及公园绿地区等，分别记为1类区、2类区、……、5类区，不同的区域环境受人类活动影响的程度不同。
现对某城市城区土壤地质环境进行调查。
为此，将所考察的城区划分为间距1公里左右的网格子区域，按照每平方公里1个采样点对表层土（0~10 厘米深度）进行取样、编号，并用GPS记录采样点的位置。
应用专门仪器测试分析，获得了每个样本所含的多种化学元素的浓度数据。
另一方面，按照2公里的间距在那些远离人群及工业活动的自然区取样，将其作为该城区表层土壤中元素的背景值。
附件1列出了采样点的位置、海拔高度及其所属功能区等信息，附件2列出了8种主要重金属元素在采样点处的浓度，附件3列出了8种主要重金属元素的背景值。
现要求你们通过数学建模来完成以下任务：(1) 给出8种主要重金属元素在该城区的空间分布，并分析该城区内不同区域重金属的污染程度。
(2) 通过数据分析，说明重金属污染的主要原因。
(3) 分析重金属污染物的传播特征，由此建立模型，确定污染源的位置。
(4) 分析你所建立模型的优缺点，为更好地研究城市地质环境的演变模式，还应收集什么信息？有了这些信息，如何建立模型解决问题？题目A题城市表层土壤重金属污染分析摘要：本文研究的是某城区警车配置及巡逻方案的制定问题，建立了求解警车巡逻方案的模型，并在满足D1的条件下给出了巡逻效果最好的方案。
在设计整个区域配置最少巡逻车辆时，本文设计了算法1：先将道路离散化成近似均匀分布的节点，相邻两个节点之间的距离约等于一分钟巡逻路程。
2011全国大学生数学建模B
sij 1 sij 0 s.t. sij 1 jJ s 1 ij iI
(cij 3km) (cij 3km) (i 1 92) ( j 1 20)
s
ij
路口由一个服务台管辖： sij 1(i I )
jJ
sij 1( j J ) 服务台管辖路口数至少为1： iI
问题一（ 2 ）的思路分析与模型建立
问题一（ 2 ） 问题的数学表达：
min f 2 max cij x ij
1i 20 1 j 13
1 ，服务台i对要道j进行封锁 xij 0 ，服务台i不对要道j进行封锁
最大时间最小：
20 xij 1, j 1 13 i 1 13 s.t. xij 1, i 1 20 j 1 x 0或1 ij
问题二（ 2 ）的思路分析与模型建立
问题二（ 2 ） 问题的数学表达：
：嫌犯在t+3内行驶的最大区域
M in T s.t. flag Qt 3 , P 1
：嫌犯在t+3内行使最大区域边界点集；
1 可以分配警力，在t时间到达Qt 3中得路口 flag Qt 3 , P 0 无法分配警力，在t时间到达Qt 3中得路口
问题二
问题二：
针对全市（主城六区 A ， B ， C ， D ， E ， F ）的具体情况，按照设置 交巡警服务平台的原则和任务，分析研究该市现有交巡警服务平台 设置方案（参见附件）的合理性，如果有明显不合理，请给出解决 方案；
如果该市地点 P （第 32 个节点）处发生了重大刑事案件，在案发 3 分钟后接到报警，犯罪嫌疑人已驾车逃跑。为了快速搜捕嫌疑犯， 请给出调度全市交巡警服务平台警力资源的最佳围堵方案。
2011高教社杯全国大学生数学建模竞赛B题(题目改变)参考答案
交巡警服务平台的设置与调度优化分析摘要本文综合应用了Floyd算法，匈牙利算法，用matlab计算出封锁全市的时间为1.2012小时。
并在下面给出了封锁计划。
为了得出封锁计划，首先根据附件2的数据将全市的道路图转为邻接矩阵，然后根据邻接矩阵采用Floyd算法计算出该城市任意两点间的最短距离。
然后从上述矩阵中找到各个交巡警平台到城市各个出口的最短距离，这个最短距离矩阵即可作为效益矩阵，然后运用匈牙利算法，得出分派矩阵。
根据分派矩阵即可制定出封锁计划：96-151,99-153,177-177,175-202,178-203,323-264,181-317, 325-325,328-328，386-332,322-362,100-387,379-418,483-483, 484-541,485-572。
除此以外，本人建议在编号为175的路口应该设置一个交巡警平台，这样可以大大减少封锁全市的时间，大约可减少50%。
关键词: Floyd算法匈牙利算法 matlab一、问题重述“有困难找警察”，是家喻户晓的一句流行语。
警察肩负着刑事执法、治安管理、交通管理、服务群众四大职能。
为了更有效地贯彻实施这些职能，需要在市区的一些交通要道和重要部位设置交巡警服务平台。
每个交巡警服务平台的职能和警力配备基本相同。
由于警务资源是有限的，如何根据城市的实际情况与需求合理地设置交巡警服务平台、分配各平台的管辖范围、调度警务资源是警务部门面临的一个实际课题。
试就某市设置交巡警服务平台的相关情况，建立数学模型分析研究下面的问题：警车的时速为60km/h, 现有突发事件，需要全市紧急封锁出入口，试求出全市所有的交巡警平台最快的封锁计划，一个出口仅需一个平台的警力即可封锁。
二、模型假设1、假设警察出警时的速度相同且不变均为60/km h 。
2、假设警察出警的地点都是平台处。
3、假设警察接到通知后同时出警，且不考虑路面交通状况。
三、符号说明及一些符号的详细解释A 存储全市图信息的邻接矩阵 D 任意两路口节点间的最短距离矩阵X 01-规划矩阵ij a ,i j 两路口节点标号之间直达的距离 ij d 从i 路口到j 路口的最短距离 ij b 从i 号平台到j 号出口的最短距离ij x 取0或1，1ij x =表示第i 号平台去封锁j 号出口在本文中经常用到,i j ，通常表示路口的编号，但是在ij d ，ij b ，ij x 不再表示这个意思，i 表示第i 个交巡警平台，交巡警平台的标号与附件中给的略有不同，如第21个交巡警平台为附件中的标号为93的交巡警平台，本文的标号是按照程序的数据读取顺序来标注的，在此声明；j 表示第j 个出口，如：第5个出口对应于附件中的路口编号为203的出口。
2011高教社杯全国大学生数学建模竞赛B题省一等奖
2011高教社杯全国大学生数学建模竞赛B题省一等奖承诺书我们仔细阅读了中国大学生数学建模竞赛的竞赛规则.我们完全明白，在竞赛开始后参赛队员不能以任何方式（包括电话、电子邮件、网上咨询等）与队外的任何人（包括指导教师）研究、讨论与赛题有关的问题。
我们知道，抄袭别人的成果是违反竞赛规则的, 如果引用别人的成果或其他公开的资料（包括网上查到的资料），必须按照规定的参考文献的表述方式在正文引用处和参考文献中明确列出。
我们郑重承诺，严格遵守竞赛规则，以保证竞赛的公正、公平性。
如有违反竞赛规则的行为，我们将受到严肃处理。
我们参赛选择的题号是（从A/B/C/D中选择一项填写）： B我们的参赛报名号为（如果赛区设置报名号的话）：B甲00226所属学校（请填写完整的全名）：参赛队员(打印并签名) ：1.2.3.指导教师或指导教师组负责人(打印并签名)：日期： 2011 年 9 月 12 日赛区评阅编号（由赛区组委会评阅前进行编号）：编号专用页赛区评阅编号（由赛区组委会评阅前进行编号）：全国统一编号（由赛区组委会送交全国前编号）：全国评阅编号（由全国组委会评阅前进行编号）：交巡警服务平台的设置与调度摘要对于给各个交巡警服务平台分配管辖范围的问题，首先运用Dijkstra算法求出A 区交通网络中的任一路口节点到其他路口节点的最短路经值，再从道路的两个节点出发，选出具离它最近的交巡警服务平台，那么此道路就由所选的服务平台来管辖，这样可以依次选出各条道路所对应的交巡警服务平台，那么各交巡警服务平台相对应的管辖范围就能划分出来。
对于调度20各服务平台来封锁13条交通要道，也即13个路口节点的情况，假设每个路口节点只需一个服务平台的警力资源来封锁，建立一个有路程约束的最佳调度方案，得出进出城区的标号为12、14、16、21、22、23、24、28、29、30、38、48、62的路口节点分别由标号为12、9、16、14、10、13、11、15、7、8、2、5、4的交巡警服务平台的警力资源来封锁。
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2011高教社杯全国大学生数学建模竞赛题目（请先阅读“全国大学生数学建模竞赛论文格式规范”）B题交巡警服务平台的设置与调度“有困难找警察”，是家喻户晓的一句流行语。
警察肩负着刑事执法、治安管理、交通管理、服务群众四大职能。
为了更有效地贯彻实施这些职能，需要在市区的一些交通要道和重要部位设置交巡警服务平台。
每个交巡警服务平台的职能和警力配备基本相同。
由于警务资源是有限的，如何根据城市的实际情况与需求合理地设置交巡警服务平台、分配各平台的管辖范围、调度警务资源是警务部门面临的一个实际课题。
试就某市设置交巡警服务平台的相关情况，建立数学模型分析研究下面的问题：（1）附件1中的附图1给出了该市中心城区A的交通网络和现有的20个交巡警服务平台的设置情况示意图，相关的数据信息见附件2。
请为各交巡警服务平台分配管辖范围，使其在所管辖的范围内出现突发事件时，尽量能在3分钟内有交巡警（警车的时速为60km/h）到达事发地。
对于重大突发事件，需要调度全区20个交巡警服务平台的警力资源，对进出该区的13条交通要道实现快速全封锁。
实际中一个平台的警力最多封锁一个路口，请给出该区交巡警服务平台警力合理的调度方案。
根据现有交巡警服务平台的工作量不均衡和有些地方出警时间过长的实际情况，拟在该区内再增加2至5个平台，请确定需要增加平台的具体个数和位置。
（2）针对全市（主城六区A，B，C，D，E，F）的具体情况，按照设置交巡警服务平台的原则和任务，分析研究该市现有交巡警服务平台设置方案（参见附件）的合理性。
如果有明显不合理，请给出解决方案。
如果该市地点P（第32个节点）处发生了重大刑事案件，在案发3分钟后接到报警，犯罪嫌疑人已驾车逃跑。
为了快速搜捕嫌疑犯，请给出调度全市交巡警服务平台警力资源的最佳围堵方案。
附件1：A区和全市六区交通网络与平台设置的示意图。
附件2：全市六区交通网络与平台设置的相关数据表（共5个工作表）。
附图1：A区的交通网络与平台设置的示意图附图2：全市六区交通网络与平台设置的示意图说明：（1）图中实线表示市区道路；红色线表示连接两个区之间的道路；（2）实圆点“·”表示交叉路口的节点，没有实圆点的交叉线为道路立体相交；（3）星号“*”表示出入城区的路口节点；（4）圆圈“○”表示现有交巡警服务平台的设置点；（5）圆圈加星号“○*”表示在出入城区的路口处设置了交巡警服务平台；（6）附图2中的不同颜色表示不同的区。
说明：A列：是全市交通网络中路口节点的标号（序号）B列：路口节点的横坐标X，是在交通网络中的实际横坐标值C列：路口节点的纵坐标Y，是在交通网络中的实际纵坐标值D列：路口节点所属的区E列：各路口节点的发案率是每个路口平均每天的发生报警案件数量地图距离和实际距离的比例是1:100000,即1毫米对应100米坐标的长度单位为毫米全市路口节点标号路口的横坐标X路口的纵坐标Y路口所属区域发案率(次数)1 413 359 A 1.72 403 343 A 2.13 383.5 351 A 2.24 381 377.5 A 1.75 339 376 A 2.16 335 383 A 2.57 317 362 A 2.48 334.5 353.5 A 2.49 333 342 A 2.110 282 325 A 1.611 247 301 A 2.612 219 316 A 2.413 225 270 A 2.214 280 292 A 2.515 290 335 A 2.116 337 328 A 2.618 432 371 A 1.919 418 374 A 1.820 444 394 A 1.921 251 277 A 1.422 234 271 A 1.423 225 265 A 2.424 212 290 A 1.125 227 300 A 1.626 256 301 A 1.227 250.5 306 A 0.828 243 328 A 1.329 246 337 A 1.430 314 367 A 2.131 315 351 A 1.632 326 355 A 1.533 327 350 A 1.434 328 342.5 A 1.735 336 339 A 1.436 336 334 A 1.137 331 335 A 0.138 371 330 A 1.239 371 333 A 1.440 388.5 330.5 A 1.741 411 327.5 A 1.442 419 344 A 1.443 411 343 A 1.744 394 346 A 1.145 342 342 A 1.446 342 348 A 1.247 325 372 A 1.648 315 374 A 1.449 342 372 A 1.250 345 382 A 1.151 348.5 380.5 A 0.852 351 377 A 0.653 348 369 A 1.454 370 363 A 0.955 371 353 A 156 354 374 A 0.557 363 382.5 A 0.858 357 387 A 1.159 351 382 A 0.960 369 388 A 0.762 381 381 A 1.263 391 375 A 1.464 392 366 A 0.865 395 361 A 0.766 398 362 A 0.867 401 359 A 0.868 405 360 A 0.969 410 355 A 1.170 408 350 A 0.971 415 351 A 1.172 418 347 A 0.873 422 354 A 0.974 418.5 356 A 1.175 405.5 364.5 A 0.876 405 368 A 1.177 409 370 A 0.878 417 364 A 0.879 420 370 A 0.880 424 372 A 0.881 438 368 A 1.482 438.5 373 A 1.183 434 376 A 0.984 438 385 A 185 440 392 A 1.286 447 392 A 1.487 448 381 A 1.188 444.5 383 A 0.989 441 385 A 1.490 440.5 381.5 A 0.991 445 380 A 0.992 444 360 A 0.893 140 130 B 1.694 145 118 B 1.695 160 96 B 1.696 142.5 71 B 2.197 150 70 B 1.898 186 145 B 1.699 158 73.5 B 2.6 100 121 68 B 2.6 101 157 145 B 1.1 102 158 138.5 B 0.9 103 159 135 B 0.5 104 133 114 B 0.7105 137.5 113 B 0.4 106 144 112 B 0.8 107 139 117 B 0.2 108 144.5 115 B 0.8 109 151 113 B 0.6 110 151.5 118 B 0.9 111 150 111 B 0.8 112 158 118 B 1.1 113 159 109 B 0.8 114 164 108.5 B 0.4 115 163 105 B 0.7 116 149 99.5 B 1.2 117 143 102 B 0.8 118 137 103 B 0.9 119 131 103 B 0.5 120 130 100 B 0.6 121 127 102 B 0.6 122 125 98 B 0.8 123 129 96 B 0.9 124 130 90 B 0.4 125 124 90 B 0.7 126 136 96 B 1.1 127 136 90 B 0.8 128 142 96 B 0.8 129 148 96 B 0.7 130 142 91 B 0.6 131 147 91 B 0.7 132 128 71 B 1.2 133 136.5 76 B 0.8 134 142 79 B 1.1 135 147 81 B 0.8 136 154 86 B 0.9 137 148.5 74.5 B 1.1 138 140 70 B 0.6 139 140 63 B 0.7 140 137.5 63 B 0.8 141 138 59 B 0.4 142 143 63 B 1.1 143 151 69 B 0.8 144 153 63 B 1.1 145 143 60 B 0.7 146 143 57 B 0.6 147 143 51.5 B 0.8 148 160 65 B 1.1149 162 59 B 0.6 150 141 49 B 0.4 151 143 40 B 0.8 152 151 44 B 0.5 153 150 33 B 0.1 154 164 124 B 0.6 155 171 125 B 0.7 156 165.5 139 B 1.1 157 181 131 B 1.4 158 176 141 B 1.6 159 170 140 B 0.8 160 168 145 B 0.6 161 166 150 B 0.8 162 176 145 B 0.6 163 180 149 B 0.7 164 183 145 B 1.1 165 202 131 B 1.1 166 137.5 462 C 2.6 167 167 399 C 2.2 168 376 400 C 1.4 169 210 390 C 2.6 170 263 445 C 2.2 171 284 409 C 1.9 172 278.5 425 C 2.2 173 295 382 C 2 174 299 444 C 2.6 175 362 443 C 2.2 176 410 408.5 C 2.1 177 395 520 C 2.2 178 277 496 C 1.7 179 235 465 C 2.2 180 200 466.5 C 1.9 181 167 462 C 2.4 182 225 443 C 2.4 183 400 447 C 1.2 184 414 422 C 1.4 185 424 400 C 1.2 186 411 396 C 1.4 187 420 401 C 0.8 188 403 404 C 1.2 189 376 406 C 0.9 190 380 404 C 0.8 191 377 424 C 0.8 192 374 424 C 0.8193 370 423 C 0.4 194 368 427.5 C 0.9 195 374 431 C 1.2 196 365 448 C 1.4 197 356 450 C 1.4 198 358 459 C 1.2 199 354 495 C 1.1 200 357 513 C 1.2 201 359 528 C 0.4 202 347 553 C 0.5 203 261 537.5 C 0.8 204 270 514 C 1.4 205 313 511.5 C 0.4 206 324 511 C 0.8 207 333 511 C 0.7 208 334 497 C 0.8 209 323 497 C 0.7 210 312 498 C 1.1 211 317 451 C 1.1 212 316 448 C 0.8 213 315.5 444 C 0.7 214 316 434 C 0.9 215 318 412 C 1.2 216 291.5 415 C 1.4 217 284 425 C 1.4 218 281 421 C 1.6 219 299 434 C 1.4 220 302 451 C 1.4 221 305 457 C 1.2 222 281 458.5 C 1.1 223 274 448 C 0.8 224 273.5 444 C 0.9 225 267 446 C 1.1 226 270 440 C 0.9 227 275 422 C 1.1 228 276 419 C 1.2 229 270 415 C 0.8 230 276 405 C 1.4 231 288 403 C 1.4 232 293.5 392.5 C 1.4 233 296 387 C 1.1 234 303 386 C 1.4 235 298.5 378 C 1.6 236 293 376 C 1.2237 296 372 C 1.7 238 276 352 C 1 239 250 350 C 1.4 240 247 384 C 1.2 241 262 399 C 1.4 242 269 397 C 1.2 243 276 402 C 1.3 244 282 398.5 C 1.2 245 282 386.5 C 1.1 246 273 389 C 1.2 247 276 361 C 1.1 248 138.5 378 C 0.8 249 155 396 C 1.2 250 163 390 C 1.5 251 173 364 C 1.2 252 183 370 C 1.2 253 238 382 C 0.7 254 213 412 C 0.8 255 189 413 C 1.1 256 210 433 C 0.9 257 201 434 C 0.9 258 150 400 C 1.2 259 135 395 C 1.2 260 143 407 C 1.1 261 142 414 C 0.8 262 140 430 C 1.1 263 121 432 C 0.9 264 109 441 C 0.5 265 138.5 442 C 1.2 266 167 442 C 1.6 267 168 435 C 1.4 268 184 440 C 1.2 269 194 442 C 0.9 270 200 442 C 1.4 271 212 443 C 1.6 272 220 443 C 1.7 273 246 444 C 2.1 274 246 455 C 1.4 275 252 458 C 1.2 276 257 460.5 C 1.5 277 255.5 466 C 1.2 278 249 464 C 1.1 279 247 469 C 0.8 280 254 472 C 0.7281 251.5 477 C 1.1 282 259 478 C 0.8 283 261 470 C 0.4 284 255 494 C 1.4 285 240 495 C 1.4 286 241 514 C 0.8 287 236 514 C 0.7 288 235 496 C 0.7 289 232 487 C 0.8 290 235.5 486.5 C 0.8 291 245 474 C 1.2 292 225 457.5 C 1.4 293 225 451 C 1.6 294 219 451 C 1.4 295 219 462 C 1.2 296 228.5 472 C 1.6 297 213 481 C 1.4 298 211 487 C 1 299 208.5 496 C 1.2 300 206 507 C 0.8 301 206 515 C 1.2 302 200 514 C 0.7 303 200 507 C 1.2 304 200 497 C 1.3 305 200 484 C 1.4 306 206 466 C 1.4 307 194 466 C 1.4 308 184 463.5 C 1.5 309 184 475 C 0.8 310 193.5 475 C 0.7 311 193 484 C 0.9 312 184 484 C 0.6 313 184 496.5 C 0.8 314 192.5 496.5 C 0.7 315 192 507 C 0.9 316 192 514 C 0.8 317 170 516.5 C 0.6 318 168 507 C 1.1 319 167 495.5 C 1.4 320 101 343 D 2.4 321 91 355 D 1.7 322 70 377 D 2.5 323 46 371 D 2.4 324 56 424 D 2.1325 20 442 D 2.2 326 74 326 D 2.6 327 76 302 D 2.1 328 15 240 D 2.6 329 28 161 D 0.4 330 34.5 164.5 D 0.1 331 30 181 D 0.6 332 27 206 D 0.2 333 42 242 D 1.4 334 30 246 D 1.6 335 31 254 D 1.1 336 39 254 D 1.2 337 50 289 D 0.7 338 72 288 D 1.1 339 60 246 D 0.7 340 95 299 D 1.4 341 81 297 D 1.6 342 80 287 D 1.4 343 67 314 D 1.7 344 21 330 D 1.1 345 36 360 D 1.2 346 76 344 D 0.8 347 97 339 D 2.4 348 103 337 D 1.2 349 104 341 D 1.1 350 97 345 D 1.6 351 89 345 D 0.8 352 81 344.5 D 0.8 353 81 350 D 0.4 354 89 350 D 0.7 355 92.5 351.5 D 1.1 356 88 353 D 1.4 357 81.5 353 D 0.9 358 87 359 D 1.1 359 84 361 D 0.9 360 76 355 D 0.8 361 58.5 370 D 0.6 362 34 306 D 0.1 363 38 418.5 D 1.4 364 61 425 D 1.4 365 57 429 D 1.6 366 60 433 D 1.4 367 85 369 D 1.9 368 107.5 362 D 1.4369 131 366.5 D 1.2 370 170 342 D 1.2 371 174 340 D 1.5 372 232.5 264 E 2.4 373 202 223 E 1.9 374 241 210 E 2.4 375 235 197.5 E 2.6 376 228 173 E 2.6 377 214 164 E 2.6 378 278 196 E 2.6 379 267 168 E 2.4 380 90 167 E 1 381 123 177.5 E 1.1 382 143 153 E 1.9 383 192 264 E 2.6 384 145 285 E 2.4 385 133 255 E 2.4 386 90 198 E 1.7 387 21 15 E 1.1 388 60 68 E 0.8 389 70 84 E 0.2 390 27 149 E 1.6 391 62 143 E 0.9 392 58 176 E 1.4 393 58 160 E 0.6 394 72 163 E 0.7 395 70 176.5 E 0.7 396 90 178 E 0.8 397 115 168 E 0.6 398 115 177.5 E 0.8 399 123 168 E 0.7 400 123 164 E 0.6 401 123 155 E 0.7 402 143 164 E 0.9 403 144 168 E 1.2 404 149 177 E 0.9 405 128 178 E 0.9 406 128 188 E 1.5 407 164 194 E 1.7 408 156 177 E 0.8 409 168 177 E 1.1 410 156 169 E 0.8 411 167 168 E 0.8 412 172 167 E 0.9413 167 164 E 0.2 414 160 164 E 0.7 415 163 153.5 E 1.2 416 186 168 E 1.6 417 269 133 E 1.6 418 295 112 E 1.1 419 302 112 E 1.4 420 316 141 E 1.6 421 278 143 E 1.7 422 284 173 E 1.4 423 257.5 170 E 1.9 424 239 198 E 0.4 425 241 198 E 0.3 426 246 199 E 0.6 427 246.5 202 E 0.4 428 240 202 E 0.4 429 236 201 E 1.1 430 231 199 E 0.1 431 232 206.5 E 0.6 432 239.5 207.5 E 0.5 433 242 206 E 0.2 434 235 209.5 E 0.4 435 237.5 212 E 0.1 436 246 208 E 0.4 437 200 194 E 1.1 438 170 222 E 1.6 439 59 189 E 0.8 440 72 189 E 0.9 441 90 187.5 E 0.6 442 74 198 E 0.7 443 60 196 E 0.4 444 90 211 E 1.6 445 151 236 E 1.4 446 160 244 E 1.5 447 90 222 E 0.8 448 129 248 E 1.7 449 142 265 E 2.1 450 152 255.5 E 1.1 451 155 258 E 0.6 452 163 258 E 0.8 453 171 258 E 1.1 454 171 252.5 E 0.4 455 171 247 E 1.2 456 214 235 E 1.1457 244 238 E 1.1 458 268 237 E 1.1 459 259 255 E 1.1 460 188 261 E 1.4 461 184 253 E 1.2 462 171 263 E 1.1 463 171 268 E 0.8 464 163 268 E 0.9 465 154 268.5 E 0.7 466 151 275 E 0.4 467 148 274 E 1.5 468 162 277.5 E 1.5 469 177 281 E 0.7 470 187 284 E 1.4 471 155 316 E 1.6 472 159 292 E 1.8 473 125 267 E 1.8 474 107 285 E 1.6 475 382.5 267 F 2.4 476 373 250 F 1.9 477 330 219 F 0.8 478 400 247 F 2.3 479 441 442 F 1.7 480 417 312 F 1.5 481 332 246 F 1.9 482 321 275 F 1.7 483 403 140 F 2.1 484 420 269 F 2.4 485 455 335 F 1.9 486 295.5 238 F 1.4 487 294 244 F 1.1 488 316 300 F 1.5 489 308 257.5 F 1.2 490 327 255 F 0.8 491 316 236 F 1.4 492 314 230 F 0.9 493 313 223 F 0.6 494 317 215 F 0.2 495 318.5 222 F 0.3 496 320 229 F 0.5 497 326.5 227.5 F 0.7 498 325 220 F 0.6 499 323 213 F 0.4 500 329 212 F 0.7501 332 226 F 0.7 502 334 210.5 F 0.6 503 346 209 F 0.7 504 342 200 F 0.8 505 356 202 F 0.7 506 358 195 F 0.6 507 345 194 F 0.4 508 348 188 F 0.4 509 357.5 188 F 0.6 510 359 159 F 1.1 511 404 161 F 1.2 512 403 202 F 0.8 513 379 202 F 0.7 514 386 213 F 0.8 515 373 213 F 0.6 516 363 212 F 0.4 517 362 218 F 0.8 518 354 216.5 F 0.6 519 348 215 F 0.9 520 349 222 F 0.7 521 353 223 F 0.8 522 371 224 F 0.8 523 371 218.5 F 0.6 524 375 219 F 0.4 525 388.5 218 F 1.1 526 405 213.5 F 0.8 527 389 224.5 F 0.9 528 388 233 F 0.6 529 353 229.5 F 0.8 530 334 232 F 0.7 531 336 239 F 1.1 532 352 247 F 1.2 533 353 236 F 0.6 534 362.5 236 F 0.8 535 370 236 F 1.1 536 388 237 F 1.2 537 395.5 237.5 F 1.4 538 395 233 F 1.1 539 408.5 227 F 1.5 540 430 237 F 1.4 541 450 268 F 0.1 542 394 254 F 1.4 543 387 250 F 0.9 544 383 250 F 1.1545 369 249.5 F 0.8 546 367.5 249 F 0.7 547 362 249 F 0.8 548 350 251 F 0.6 549 348 255 F 1.4 550 355 265 F 1.1 551 367 265 F 0.8 552 367 257.5 F 1.2 553 375 258 F 1.4 554 376 260 F 1.1 555 381 260 F 1.7 556 378 266 F 1.4 557 380 270.5 F 1.2 558 371 284 F 1.1 559 356.5 281 F 1.4 560 338 297 F 1.2 561 372 307 F 1.4 562 398 308 F 1.5 563 392 277 F 1.1 564 382.5 276 F 0.9 565 396 270 F 1.4 566 411 291 F 1.2 567 424 297 F 0.8 568 435 319 F 0.9 569 434 307 F 0.7 570 430 295 F 1.4 571 441 309 F 1.2 572 470 342 F 0.2 573 468 432 F 1.2 574 455 361 F 0.6 575 453 400 F 0.6 576 425 433 F 0.8 577 462 437 F 1.4 578 481 457 F 0.6 579 462 447 F 1.2 580 440 449 F 1.4 581 423 448 F 1 582 435 507.5 F 0.4。