加勒比久久综合,国产精品伦一区二区,66精品视频在线观看,一区二区电影

合肥生活安徽新聞合肥交通合肥房產生活服務合肥教育合肥招聘合肥旅游文化藝術合肥美食合肥地圖合肥社保合肥醫院企業服務合肥法律

代做Spatial Networks for Locations

時間:2024-02-16  來源:合肥網hfw.cc  作者:hfw.cc 我要糾錯



Background
Spatial Networks for Locations
 Locations are connected via roads (we assume traders can travel in both
directions!)  These locations form a spatial network.  As traders used horses for travelling, they couldn’t travel too far!
Pottery Trade
Pottery trade was very active at that times. Each location had its own supply and demandfor pottery. The supply and demand were communicated by traders who also formed their
own networks. They also potentially communicated the prices, but in these project wewill
disregard this information.
Social Networks for Traders
Traders living in some locations know each other and exchange information about supplyand demand via postal services. These traders for a social network.
How to Represent Networks
Each network can be presented as a graph. In this project, we will focus on undirectedgraphs: both social and spatial networks can be represented as graphs:
1. Spatial networks: nodes correspond to locations, and edges —to roads betweenthem (both directions)
2. Social networks: nodes correspond to traders, and edges connect those who
know each other (communicate)
Networks/graphs can be very different!
Project Questions
1. Represent road maps and trader networks as graphs
2. Find the shortest path between any two locations (return the shortest path andthedistance)
3. (Static traders) Find the best trading options for a particular trader residing in aparticular location. Core concepts: Itineraries
Itineraries provide the basis for our spatial network. They are provided as a list of (L1,L2, distance) tuples; listed in any order. L1 and L2 are provided as strings, distance is an integer number (miles).
In the example:
>>> itineraries = [('L1', 'L2', 20), ('L2', 'L3', 10), ('L1', 'L4', 15), ('L4','L5',5), ('L4', 'L8', 20), ('L5', 'L8', 22), ('L5', 'L6', 6), ('L6', 'L7', 20)]
Supply and Demand of Goods (Pottery)
Each location has its own supply and demand in pottery: supply is provided as a positivenumber, demand — as a negative. Locations with the highest demand should be servedfirst. Assume both numbers are integers. This is provided as a dictionary (in no particular order)
>>> status = {'L1':50, 'L2':-5, 'L4':-40, 'L3':5, 'L5':5, 'L8':10, 'L6':10, 'L7':-30}Trader Locations
Traders reside in some but not all locations. Only locations where traders are present cantrade. Each location can have maximum a single trader. Traders are provided as strings.
Trader locations are provided as a dictionary (in no particular order). In the example:
>>> trader_locations = {'T1':'L1', 'T2': 'L3', 'T3':'L4', 'T4':'L8', 'T5':'L7','T6':'L5'}
Social network of Traders
Traders also form a social network. A trader only trades within their own network
(considers friends only). Traders also have access to supplies and demands in the
corresponding locations. Trader friendships are provided as a list of tuples (in no particular order):
>>> traders = [('T1','T2'), ('T2', 'T5'), ('T3', 'T1'), ('T3', 'T5'), ('T3', 'T6')]Q1
Write a function create_spatial_network(itineraries) that takes itineraries (a list of
tuples) and returns for each location its neighbors and distances to them. A location is
considered to be a neighbour of another location if it can be reached by a single road (oneedge).
Input:
**3; itineraries: a list of tuples, where each tuple is of the
form (location1, location2, distance). location1 and location2 are the stringlabels for these locations and distance is an integer. Your function should return a list of tuples, where each tuple is of the
form (location, neighbours). neighbours should be of the
form [(neighbour1, distance1), (neighbour2, distance2), ...] and be sorted by their
distances (in the increasing order). If two or more neighbors have the same distance tothe location, tie-break by alphanumeric order on their labels. Note that in addition to the neighbors, the overall list has to be sorted. You may assume: **3; Distances are non-negative integer values
**3; Inputs are correctly formatted data structures and types
**3; There are no duplicate entries itineraries, and in each neighbor pair only appear
once (i.e. no [('L1', 'L2', 20), ('L2', 'L1', 20)])
Here is a diagram of an example network:
For the network above, this would be a possible itineraries and the function should
return the following:
>>> itineraries = [('L1', 'L2', 20), ('L2', 'L3', 10), ('L1', 'L4', 15), ('L4','L5',5), ('L4', 'L8', 20), ('L5', 'L8', 22), ('L5', 'L6', 6), ('L6', 'L7', 20)]
>>> create_spatial_network(itineraries)
[('L1', [('L4', 15), ('L2', 20)]), ('L2', [('L3', 10), ('L1', 20)]), ('L3', [('L2',10)]),('L4', [('L5', 5), ('L1', 15), ('L8', 20)]), ('L5', [('L4', 5), ('L6', 6), ('L8', 22)]),('L6', [('L5', 6), ('L7', 20)]), ('L7', [('L6', 20)]), ('L8', [('L4', 20), ('L5', 22)])]A different example (not pictured):
>>> itineraries = [('L4', 'L1', 2), ('L3', 'L1', 5), ('L1', 'L5', 5), ('L2', 'L5',1)]>>> create_spatial_network(itineraries)
[('L1', [('L4', 2), ('L3', 5), ('L5', 5)]), ('L2', [('L5', 1)]), ('L3', [('L1',5)]),('L4', [('L1', 2)]), ('L5', [('L2', 1), ('L1', 5)])]
Q2
Write a function sort_demand_supply(status) that takes a dictionary of demands andsupplies and returns the information as a list of tuples sorted by the value so that locationswith greatest demands (the most negative number) are provided first.
Input: **3; status: a dictionary of demands and supplies. The keys are the location labels
(strings) and the values are integers, where a positive value represents supply
and a negative value represents demand. Your function should return a list of tuples, where each tuple is of the
form (location, demand_supply), and the list should be sorted in ascending order by
their demand_supply (i.e. greatest demand to greatest supply). If two or more locationshave the same demand or supply, tie-break by alphanumeric order on their labels. You may assume: **3; Inputs are correctly formatted data structures and types
>>> status = {'L1':50, 'L2':-5, 'L4':-40, 'L3':5, 'L5':5, 'L8':10, 'L6':10, 'L7':-30}>>> sort_demand_supply(status)
[('L4', -40), ('L7', -30), ('L2', -5), ('L3', 5), ('L5', 5), ('L6', 10), ('L8',10),('L1', 50)]
Another example:
>>> status = {'L1':30, 'L2':-20, 'L4':100, 'L3':-50, 'L5':-60}
>>> sort_demand_supply(status)
[('L5', -60), ('L3', -50), ('L2', -20), ('L1', 30), ('L4', 100)]
Q3
Write a function create_social_network(traders) that takes traders, a list of tuples
specifing trader connections (edges in the trader social network) and returns a list
containing (trader, direct_connections) for each trader in traders.
Input: **3; traders: a list of tuples specifing trader connections (edges in the trader social
network). Each tuple is of the
form (trader1, trader2) where trader1 and trader2 are string names of
each trader.
Your function should return list of tuples in alphanumeric order of trader name, where
each tuple is of the form (trader, direct_connections), and direct_connections is analphanumerically sorted list of that trader's direct connections (i.e. there exists an edgebetween them in the trader social network). You may assume: **3; Inputs are correctly formatted data structures and types. Just like Q1a, you don't
need to guard against something like [('T1', 'T2'), ('T2', 'T1')] or duplicate
entries.
The pictured example:
>>> traders = [('T1','T2'), ('T2', 'T5'), ('T3', 'T1'), ('T3', 'T5'), ('T3', 'T6')]>>> create_social_network(traders)
[('T1', ['T2', 'T3']), ('T2', ['T1', 'T5']), ('T3', ['T1', 'T5', 'T6']), ('T5', ['T2','T3']),('T6', ['T3'])]
Another example (not pictured):
>>> traders = [('T1', 'T5'), ('T2', 'T6'), ('T3', 'T7'), ('T4', 'T8'), ('T1', 'T6'),('T2', 'T7'), ('T3', 'T8'), ('T4', 'T5'), ('T1', 'T7'), ('T2', 'T8'), ('T3', 'T5'),('T4','T6')]
>>> create_social_network(traders)
[('T1', ['T5', 'T6', 'T7']), ('T2', ['T6', 'T7', 'T8']), ('T3', ['T5', 'T7', 'T8']),('T4', ['T5', 'T6', 'T8']), ('T5', ['T1', 'T3', 'T4']), ('T6', ['T1', 'T2', 'T4']),('T7',['T1', 'T2', 'T3']), ('T8', ['T2', 'T3', 'T4'])]
Q4
Write a function shortest_path(spatial_network, source, target, max_bound) that
takes a spatial network, initial (source) location, target location and the maximumdistance(that a trader located in the initial location can travel) as its input and returns a tuple withashortest path and its total distance.
Input:  spatial_network: a list of tuples, where each tuple is of the
form (location, neighbours) and neighbours is of the
form [(neighbour1, distance1), (neighbour2, distance2), ...]. This
corresponds with the output of the function you wrote for Q1a.  source: the location label (string) of the initial location. **3; target: the location label (string) of the target location. **3; max_bound: an integer (or None) that specifies the maximum total distance that
your trader can travel. If max_bound is None then always return the path withminimum distance. Your function should return a tuple (path, total_distance), where path is a string of
each location label in the path separated by a - hyphen character, and total_distanceisthe total of the distances along the path.
If there's two paths with the same minimum total distance, choose the path with morelocations on it. If there's two paths with the same minimum total distance and they havethe same number of locations on the path then choose alphanumerically smaller pathstring.
If there is no path with a total distance within the max_bound then your function shouldreturn (None, None). You may assume:
 Inputs are correctly formatted data structures and types. **3; Distances are non-negative integer values. **3; The network is connected, so a path always exists, although it may not have atotal distance within the maximum bound.
>>> spatial_network = [('L1', [('L4', 15), ('L2', 20)]), ('L2', [('L3', 10), ('L1',20)]),('L3', [('L2', 10)]), ('L4', [('L5', 5), ('L1', 15), ('L8', 20)]), ('L5', [('L4',5),('L6', 6), ('L8', 22)]), ('L6', [('L5', 6), ('L7', 20)]), ('L7', [('L6', 20)]), ('L8',[('L4', 20), ('L5', 22)])]
>>> shortest_path(spatial_network, 'L1', 'L3', 50)
('L**L2-L3', 30)
>>> shortest_path(spatial_network, 'L1', 'L3', 0)
(None, None)
>>> shortest_path(spatial_network, 'L1', 'L3', 10)
(None, None)
>>> shortest_path(spatial_network, 'L1', 'L3', None)
('L**L2-L3', 30)
Q5
In this question you will be writing a
function trade(spatial_network, status_sorted, trader_locations, trader_network, max_dist_per_unit=3) that makes a single trade.
Input:
**3; spatial_network: a list of tuples, where each tuple is of the
form (location, neighbours) and neighbours is of the
form [(neighbour1, distance1), (neighbour2, distance2), ...]. This
corresponds with the output of the function you wrote for Q1a. **3; status_sorted: a list of tuples, where each tuple is of the
form (location, demand_supply), and the list is sorted in ascending order by
their demand_supply (i.e. greatest demand to greatest supply) with ties brokenalphanumerically on location label. This corresponds with the output of the
function you wrote for Q1b. **3; trader_locations: a dictionary of trader locations. The structure of this
is trader_name: trader_location, where
both trader_name and trader_location are strings. **3; trader_network: a list of tuples in alphanumeric order of trader name, whereeach tuple is of the form (trader, direct_connections), and direct_connections is an alphanumerically sorted list of that trader's direct
connections (i.e. there exists an edge between them in the trader social network). This corresponds with the output of the function you wrote for Q1c. **3; max_dist_per_unit: a float or integer value that represents the maximumthetrader is willing to travel per unit. This parameter should have a default of 3in your
function. Your function should return a single trade as a
tuple (supplier_location, consumer_location, amount) where supplier_locationand consumer_location are location labels (strings) and amount is a positive integer. If notrade is possible return (None, None, None).
Traders from the locations with highest demand contact their social network asking for
help. Then they choose the contacts worth travelling to, based on distance and the
amount of supply there. The trade shoud be determined as follows:
1. Find the location with the highest demand, this will be the consumer location. 2. Find the trader at the consumer location (skip this location and go back to step1if
there are no traders at this location) and consider the trader's connections. 3. A supplier location can only supply to the consumer location if their status is
positive (i.e. they have items to supply) and can supply an amount up to this value(i.e. they can't supply so much that they result in having a demand for the itemthey are supplying). 4. If a supplier location is directly neighbouring by a single road (adjacent) to theconsumer location then the distance used is the direct distance between the twolocations, even if there exists a shorter route via other locations. If the supplier andconsumer are not adjacent then the shortest_path function should be used todetermine the distance. 5. The trader will trade with the connection that has the highest amount of units tosupply, subject to meeting the max_dist_per_unit of the distance/units ratio. 6. Then if no trade is possible in this location, consider the next location. Return (None, None, None) if all locations have been considered. You may assume: **3; Inputs are correctly formatted data structures and types. **3; Distances are non-negative integer values. **3; There will be at most one trader at any particular location.
Consider the spatial and trader network in the image above. With a
default max_dist_per_unit of 3, the trader will only consider travelling maximum3 milesfor each unit (one direction), i.e. they will agree to travel 6 miles for get 2 pottery units but
not a single one.
In the example, we have 'L4' as the location with the highest demand of 40 units
(demand_supply=-40) and the trader 'T3' who resides there. 'T3''s direct connectionsare ['T1', 'T5', 'T6']. We can't trade with 'T5' because at their location ('L7') there is
also demand for the items. We compare the units able to be supplied and the distance-units ratio for each potential
supplier: **3; T1:
o location: L1
o supply max: 50
o distance: 15
o so they could supply all 40 units that are demanded at L4
o distance/units = 15/40 = 0.375
**3; T6:
o location: L5
o supply max: 5
o distance: 5
o so they could supply 5 of the units that are demanded at L4
o distance/units = 5/5 = 1.0
Since T1 has the largest amount of units able to be supplied, and the distance/units ratiois below the maximum (3), this trade goes ahead and the function would
return ('L1', 'L4', 40). >>> spatial_network = [('L1', [('L4', 15), ('L2', 20)]), ('L2', [('L3', 10), ('L1',20)]),('L3', [('L2', 10)]), ('L4', [('L5', 5), ('L1', 15), ('L8', 20)]), ('L5', [('L4',5),('L6', 6), ('L8', 22)]), ('L6', [('L5', 6), ('L7', 20)]), ('L7', [('L6', 20)]), ('L8',[('L4', 20), ('L5', 22)])]
>>> status_sorted = [('L4', -40), ('L7', -30), ('L2', -5), ('L3', 5), ('L5', 5), ('L6',10), ('L8', 10), ('L1', 50)]
>>> trader_locations = {'T1':'L1', 'T2': 'L3', 'T3':'L4', 'T4':'L8', 'T5':'L7','T6':'L5'}
>>> trader_network = [('T1', ['T2', 'T3']), ('T2', ['T1', 'T5']), ('T3', ['T1','T5','T6']), ('T5', ['T2', 'T3']),('T6', ['T3'])]
>>> trade(spatial_network, status_sorted, trader_locations, trader_network)
('L1', 'L4', 40)
More examples:
>>> spatial_network = [('L1', [('L4', 2), ('L3', 5), ('L5', 5)]), ('L2', [('L5',1)]),('L3', [('L1', 5)]), ('L4', [('L1', 2)]), ('L5', [('L2', 1), ('L1', 5)])]
>>> status = {'L1':30, 'L2':-20, 'L4':100, 'L3':-50, 'L5':-60}
>>> status_sorted = [('L5', -60), ('L3', -50), ('L2', -20), ('L1', 30), ('L4',100)]>>> trader_locations = {'T1': 'L1', 'T2': 'L2'}
>>> trader_network = [('T1', ['T2']), ('T2', ['T1'])]
>>> trade(spatial_network, status_sorted, trader_locations, trader_network)
('L1', 'L2', 20)
>>> trade(spatial_network, status_sorted, trader_locations, trader_network,
max_dist_per_unit=0.001)
(None, None, None)
Q6
In this part you'll be using the trade() function from part 3a iteratively to determine thestatus after several trades. Write a
function trade_iteratively(num_iter, spatial_network, status, trader_locations, trader_network, max_dist_per_unit=3) that takes the number of iterations to perform,
the spatial network, status dictionary, trader locations dictionary, trader network, and
maximum distance per unit and returns a tuple containing the sorted status list
after num_iter trades along with a list of trades performed.
Input: **3; num_iter: the number of iterations to perform as an integer or None if the
iteration should continue until no further trades can be made. **3; spatial_network: a list of tuples, where each tuple is of the
form (location, neighbours) and neighbours is of the
form [(neighbour1, distance1), (neighbour2, distance2), ...]. This
corresponds with the output of the function you wrote for Q1a. **3; status: a dictionary of demands and supplies. The keys are the location labels
(strings) and the values are integers, where a positive value represents supply
and a negative value represents demand. **3; trader_locations: a dictionary of trader locations. The structure of this
is trader_name: trader_location, where
both trader_name and trader_location are strings. **3; trader_network: a list of tuples in alphanumeric order of trader name, whereeach tuple is of the form (trader, direct_connections), and direct_connections is an alphanumerically sorted list of that trader's direct
connections (i.e. there exists an edge between them in the trader social network). This corresponds with the output of the function you wrote for Q1c.
**3; max_dist_per_unit: a float or integer value that represents the maximumthetrader is willing to travel per unit. This parameter should have a default of 3in your
function. At each iteration, the next trade to be performed is determined by the process in part 3a. We strongly suggest using the provided trade() function to find this trade. Your functionshould update the status dictionary at each iteration. Your function should return a tuple (final_supply_sorted, trades) containing the sorteddemand-supply status after num_iter trades along with a list of trades performed. The final_supply_sorted should be a list of tuples, where each tuple is of the
form (location, demand_supply), and the list should be sorted in ascending order by
their demand_supply (i.e. greatest demand to greatest supply). If two or more locationshave the same demand or supply, tie-break by alphanumeric order on their
labels. trades should be a list of each trade performed, where a trade is of the
form (supplier_location, consumer_location, amount) where supplier_locationandconsumer_location are location labels (strings) and amount is a positive integer. You may assume: Inputs are correctly formatted data structures and types. **3; Distances are non-negative integer values.  There will be at most one trader at any particular location.
In the example pictured, only one trade can occur:
>>> spatial_network = [('L1', [('L4', 15), ('L2', 20)]), ('L2', [('L3', 10), ('L1',20)]),('L3', [('L2', 10)]), ('L4', [('L5', 5), ('L1', 15), ('L8', 20)]), ('L5', [('L4',5),('L6', 6), ('L8', 22)]), ('L6', [('L5', 6), ('L7', 20)]), ('L7', [('L6', 20)]), ('L8',[('L4', 20), ('L5', 22)])]
>>> status = {'L1': 50, 'L2': -5, 'L4': -40, 'L3': 5, 'L5': 5, 'L8': 10, 'L6': 10,'L7':-30}
>>> trader_locations = {'T1': 'L1', 'T2': 'L3', 'T3': 'L4', 'T4': 'L8', 'T5': 'L7','T6':'L5'}
>>> trader_network = [('T1', ['T2', 'T3']), ('T2', ['T1', 'T5']), ('T3', ['T1','T5','T6']), ('T5', ['T2', 'T3']),('T6', ['T3'])]
>>> trade_iteratively(1, spatial_network, status, trader_locations, trader_network)([('L7', -30), ('L2', -5), ('L4', 0), ('L3', 5), ('L5', 5), ('L1', 10), ('L6', 10),('L8',10)], [('L1', 'L4', 40)])
>>> trade_iteratively(None, spatial_network, status, trader_locations, trader_network)([('L7', -30), ('L2', -5), ('L4', 0), ('L3', 5), ('L5', 5), ('L1', 10), ('L6', 10),('L8',10)], [('L1', 'L4', 40)])

請加QQ:99515681  郵箱:99515681@q.com   WX:codehelp 

掃一掃在手機打開當前頁
  • 上一篇:代寫ECE438、代做C/C++編程語言
  • 下一篇: cs400編程代寫、A03.FirstGit程序語言代做
  • 無相關信息
    合肥生活資訊

    合肥圖文信息
    2025年10月份更新拼多多改銷助手小象助手多多出評軟件
    2025年10月份更新拼多多改銷助手小象助手多
    有限元分析 CAE仿真分析服務-企業/產品研發/客戶要求/設計優化
    有限元分析 CAE仿真分析服務-企業/產品研發
    急尋熱仿真分析?代做熱仿真服務+熱設計優化
    急尋熱仿真分析?代做熱仿真服務+熱設計優化
    出評 開團工具
    出評 開團工具
    挖掘機濾芯提升發動機性能
    挖掘機濾芯提升發動機性能
    海信羅馬假日洗衣機亮相AWE  復古美學與現代科技完美結合
    海信羅馬假日洗衣機亮相AWE 復古美學與現代
    合肥機場巴士4號線
    合肥機場巴士4號線
    合肥機場巴士3號線
    合肥機場巴士3號線
  • 短信驗證碼 目錄網 排行網

    關于我們 | 打賞支持 | 廣告服務 | 聯系我們 | 網站地圖 | 免責聲明 | 幫助中心 | 友情鏈接 |

    Copyright © 2025 hfw.cc Inc. All Rights Reserved. 合肥網 版權所有
    ICP備06013414號-3 公安備 42010502001045

    久久99国内| 在线亚洲观看| 欧美激情视频一区二区三区免费| 裸体一区二区| 999在线精品| 综合天堂av久久久久久久| 热三久草你在线| 亚洲国产日韩欧美在线| 中文字幕av一区二区三区四区| 欧美在线高清| 不卡专区在线| 亚洲精品中文字幕乱码| 蜜桃在线一区| 国产精品一区免费在线 | 亚洲综合日本| 麻豆成人入口| 亚洲色图丝袜| 中文字幕一区二区三三| 美女网站视频一区| 97精品视频在线看| 中文亚洲欧美| 欧美日韩激情| 中日韩免视频上线全都免费| 精品久久免费| 亚州av一区| 日本不卡一区二区| 不卡亚洲精品| 校园春色亚洲| www.51av欧美视频| 亚洲欧美日韩精品一区二区| 国语产色综合| 精品久久久久中文字幕小说| 日韩电影在线观看电影| 综合一区二区三区| 日韩精品视频网站| 青草综合视频| 亚洲黄色免费看| 亚洲欧美高清| 香蕉久久a毛片| 一本色道久久综合亚洲精品不卡| 欧美+亚洲+精品+三区| 91精品国产乱码久久久久久久| 91国内精品| 国产一级成人av| 亚洲欧洲国产精品一区| 日韩欧美中文字幕一区二区三区| 亚洲图区在线| 天堂久久av| 99re8这里有精品热视频8在线| 日韩欧美一级| 日韩电影不卡一区| 日韩av电影一区| 日韩av一二三| 亚洲成人黄色| 成人中文在线| 91精品精品| 欧美成人综合| 99视频精品| 石原莉奈在线亚洲三区| 蜜桃一区二区三区在线| 欧美3p视频| av日韩中文| a∨色狠狠一区二区三区| 久久精品国产亚洲高清剧情介绍| 美女网站一区二区| 欧美日韩亚洲三区| 成人午夜888| 日韩欧美激情电影| 欧美久久精品| 亚洲国产不卡| 免费高清不卡av| 日韩在线观看不卡| 日本不卡在线视频| 亚欧洲精品视频在线观看| 久久久久九九精品影院| 成人羞羞视频播放网站| 好吊视频一区二区三区四区| 日韩中文字幕区一区有砖一区 | 麻豆视频在线看| 国产超碰精品| 欧美日韩亚洲国产精品| 日韩欧美久久| 一区三区在线欧| 石原莉奈在线亚洲三区| 女人高潮被爽到呻吟在线观看 | 色琪琪久久se色| 深夜成人福利| 亚洲欧洲中文字幕| 欧美视频二区| 亚洲免费黄色| a∨色狠狠一区二区三区| 亚洲精品黄色| 9l亚洲国产成人精品一区二三| 欧美69wwwcom| 国产精品精品国产一区二区| 久久狠狠亚洲综合| 久久国产精品免费一区二区三区| 欧美亚洲国产一区| 免费一级片91| 国产精品久久久久久久久久妞妞| 精品国产亚洲一区二区三区| 自拍欧美一区| 中文另类视频| 亚州av日韩av| 一区二区视频欧美| 丁香婷婷久久| 日韩激情欧美| 亚洲欧美bt| 日本美女一区二区三区视频| 99a精品视频在线观看| 性一交一乱一区二区洋洋av| 国产日韩一区二区三区在线播放| 亚洲图色一区二区三区| 老司机午夜免费精品视频| 日本中文字幕一区| 韩国女主播一区二区三区| 天使萌一区二区三区免费观看| 一区二区毛片| 91亚洲无吗| 国内激情视频在线观看| www.久久99| 亚洲天堂偷拍| 99久久伊人| 日本a级不卡| 91亚洲人成网污www| 亚洲日产av中文字幕| 91久久综合| 久久这里只有| 欧美日韩亚洲在线观看| 亚洲精品乱码日韩| 精品深夜福利视频| 亚洲优女在线| 在线视频亚洲欧美中文| 最近高清中文在线字幕在线观看1| 国产欧美日韩在线一区二区| 国产亚洲激情| 国产精品久久久久久av公交车 | 在线观看国产精品入口| 久久精品国产99国产精品| 久久精品1区| 精品日韩视频| 丝袜av一区| 欧美伊人久久| 亚洲第一区色| 麻豆成人久久精品二区三区小说 | 日本强好片久久久久久aaa| 日本欧美肥老太交大片| 日韩av网站在线免费观看| 四季av在线一区二区三区| 日韩成人av影视| 青青青免费在线视频| 亚洲国产中文在线二区三区免| 日韩在线不卡| 久久精品国内一区二区三区水蜜桃| 国产精品xxx| 亚洲无线视频| 亚洲色图欧美| 亚洲在线电影| 日韩电影在线一区二区三区| 在线一区av| 欧美日韩一二三四| 亚洲欧美综合久久久| 日韩有码一区二区三区| 日韩av黄色在线| 欧美日韩尤物久久| 美女少妇全过程你懂的久久| 国产精品igao视频网网址不卡日韩 | 女人av一区| 欧美三级一区| 日韩深夜视频| 亚洲网站视频| 国产精品片aa在线观看| 日韩久久久久| 亚洲国产专区| 亚洲va久久久噜噜噜久久| 婷婷综合六月| 午夜精品偷拍| 警花av一区二区三区| 久久天天久久| 三级亚洲高清视频| 女同一区二区三区| 91麻豆精品| 天天综合网天天| 在线一级成人| 日韩中文在线| 青青青爽久久午夜综合久久午夜| 狠狠色丁香久久综合频道| 一区二区三区视频播放| 欧美亚洲在线| 麻豆视频在线观看免费网站黄| 激情婷婷久久| 日本一不卡视频| 日本不卡一区二区三区高清视频| 蜜臀av性久久久久蜜臀aⅴ流畅| 久久久久久久久久久妇女| 亚州国产精品| 久久在线精品| 精品极品在线| 玖玖玖国产精品| 激情久久综合|