首页
网站开发
桌面应用
管理软件
微信开发
App开发
嵌入式软件
工具软件
数据采集与分析
其他
首页
>
> 详细
讲解C++程序设计、辅导Algorithms编程、c++编程辅导 辅导留学生Prolog|辅导Python程序
项目预算:
开发周期:
发布时间:
要求地区:
C++ Implementation of Graph Algorithms (worth 10%, due June
6th 23:59PM, late submissions not accepted)
Mingyu Guo
1 Task Description
You are asked to use C++ to solve the following puzzle.
Hint: All it takes is an algorithm mentioned in this course (with a slight twist).
Update on May 10th: The graph is undirected!
2 Submission Guideline
You must follow this guideline! Your submission will be marked automatically. Failure to
follow this guideline will result in 0.
Your submission should contain exactly one file: main.cpp.
You do not need to submit a design.
3 Puzzle
You need to redesign the road system of an imaginary country.
The country is composed of N cities (for simplicity numbered from 0 to N − 1). Some pairs of cities are
connected by bidirectional roads. We say that there is a path between different cities A and B if there exists
a sequence of unique cities C1, C2, . . . , CM, such that C1 = A and CM = B and for each index i < M, there
is a road between cities Ci and Ci+1.
The current state of the road network is miserable. Some pairs of cities are not connected by any path. On
the other hand, other pairs of cities are connected by multiple different paths, and that leads to complicated
traffic routing. You want to build some new roads and destroy some of the already existing roads in the
country so that after the reconstruction there will exist exactly one path between every pair of distinct cities.
As building new roads and destroying old ones costs a lot of money, you want to minimize the total cost
spent on the reconstruction.
You are given three two-dimensional arrays:
• country[i][j]=1 or 0: there is an existing road between city i and j if and only if country[i][j]=1.
• build[i][j]: the cost for building a road between i and j. The values of build[i][j] are represented
using English letters. A, B, . . . , Z represent 0, 1, . . . , 25 and a, b, . . . , z represent 26, 27, . . . , 51. For
example, if build[2][4]=b, then that means the cost for building a road between city 2 and city 4 is
27.
• destroy[i][j]: the cost for destroying a road between i and j. Again, the values are represented
using English letters like the above.
Your task is to find and print the minimal cost needed for the road network reconstruction.
You don’t need to worry about invalid inputs.
1
• Sample input 1: 000,000,000 ABD,BAC,DCA ABD,BAC,DCA
Note: 000,000,000 describes the two-dimensional array country. ABD,BAC,DCA describes the twodimensional
array build. ABD,BAC,DCA describes the two-dimensional array destroy. The input
format is: three strings separated by spaces; each string contains N parts separated by commas; each
part contains N characters.
Sample output 1: 3
Comment: There are three cities, totally disconnected.
• Sample input 2: 011,101,110 ABD,BAC,DCA ABD,BAC,DCA
Sample output 2: 1
Comment: Now the three cities form a connected triangle and we need to destroy one road. Optimal
solution is to destroy the road between the cities 0-1 (cost 1).
• Sample input 3: (note: all inputs are on the same line. I just couldn’t fit them in one line in this pdf.)
011000,101000,110000,000011,000101,000110
ABDFFF,BACFFF,DCAFFF,FFFABD,FFFBAC,FFFDCA
ABDFFF,BACFFF,DCAFFF,FFFABD,FFFBAC,FFFDCA
Sample output 3: 7
Comment: We have six cities forming two separate triangles. Destroy one road in each triangle (costs
1 for each road) and then join the triangles by a new road (costs 5).
• Sample input 4: 0 A A
Sample output 4: 0
Comment: One city is okay just as it is.
• Sample input 5: 0001,0001,0001,1110 AfOj,fAcC,OcAP,jCPA AWFH,WAxU,FxAV,HUVA
Sample output 5: 0
Comment: We have four cities, which are connected in such a way that there is exactly one path
between each two cities.
Thus there is nothing to reconstruct.
4 Marking
Marking will be done automatically. The total mark is 10 (1 for compiling and 9 for 9 test cases).
5 SVN Instructions
First of all, you need to create a directory under version control:
svn mkdir --parents -m "Creating ADSA Assignment 4 folder" https://version-control.adelaide.edu.au/svn/aXXXXXXX/2021/s1/adsa/assignment4/
aXXXXXXX should be your student ID. The directory path needs to be exactly “2021/s1/adsa/assignmentK”,
where “K” is the assignment number. To check out a working copy, type
svn checkout https://version-control.adelaide.edu.au/svn/aXXXXXXX/2021/s1/adsa/assignment4/ adsa-21-s1-assignment4/
cd adsa-21-s1-assignment4
svn add main.cpp
Commit the files to SVN:
svn commit -m "Adding ADSA assignment 4 main.cpp"
SVN helps keeping track of file changes (over different commits). You should commit your work early and
often.
2
6 Websubmission
You are asked to submit via the web interface https://cs.adelaide.edu.au/services/websubmission/.
The submission steps should be self-explanatory. Simply choose the correct semester, course, and assignment.
The websubmission system will automatically fetch the latest version of your work from your SVN repository
(you may also choose to submit older versions). Once your work is submitted, the system will launch a
script checking the format of your submission. Click “View Feedback” to view the results. Your mark will
be calculated offline after the deadline. You are welcome to resubmit for as many times as you wish (before
the deadline).
We will compile your code using g++ -o main.out -std=c++11 -O2 -Wall main.cpp. It is your responsibility
to ensure that your code compiles on the university system.
1
1g++ has too many versions, so being able to compile on your laptop does not guarantee that it compiles on the university
system. You are encouraged to debug your code on a lab computer (or use SSH).
3
软件开发、广告设计客服
QQ:99515681
邮箱:99515681@qq.com
工作时间:8:00-23:00
微信:codinghelp
热点项目
更多
urba6006代写、java/c++编程语...
2024-12-26
代做program、代写python编程语...
2024-12-26
代写dts207tc、sql编程语言代做
2024-12-25
cs209a代做、java程序设计代写
2024-12-25
cs305程序代做、代写python程序...
2024-12-25
代写csc1001、代做python设计程...
2024-12-24
代写practice test preparatio...
2024-12-24
代写bre2031 – environmental...
2024-12-24
代写ece5550: applied kalman ...
2024-12-24
代做conmgnt 7049 – measurem...
2024-12-24
代写ece3700j introduction to...
2024-12-24
代做adad9311 designing the e...
2024-12-24
代做comp5618 - applied cyber...
2024-12-24
热点标签
mktg2509
csci 2600
38170
lng302
csse3010
phas3226
77938
arch1162
engn4536/engn6536
acx5903
comp151101
phl245
cse12
comp9312
stat3016/6016
phas0038
comp2140
6qqmb312
xjco3011
rest0005
ematm0051
5qqmn219
lubs5062m
eee8155
cege0100
eap033
artd1109
mat246
etc3430
ecmm462
mis102
inft6800
ddes9903
comp6521
comp9517
comp3331/9331
comp4337
comp6008
comp9414
bu.231.790.81
man00150m
csb352h
math1041
eengm4100
isys1002
08
6057cem
mktg3504
mthm036
mtrx1701
mth3241
eeee3086
cmp-7038b
cmp-7000a
ints4010
econ2151
infs5710
fins5516
fin3309
fins5510
gsoe9340
math2007
math2036
soee5010
mark3088
infs3605
elec9714
comp2271
ma214
comp2211
infs3604
600426
sit254
acct3091
bbt405
msin0116
com107/com113
mark5826
sit120
comp9021
eco2101
eeen40700
cs253
ece3114
ecmm447
chns3000
math377
itd102
comp9444
comp(2041|9044)
econ0060
econ7230
mgt001371
ecs-323
cs6250
mgdi60012
mdia2012
comm221001
comm5000
ma1008
engl642
econ241
com333
math367
mis201
nbs-7041x
meek16104
econ2003
comm1190
mbas902
comp-1027
dpst1091
comp7315
eppd1033
m06
ee3025
msci231
bb113/bbs1063
fc709
comp3425
comp9417
econ42915
cb9101
math1102e
chme0017
fc307
mkt60104
5522usst
litr1-uc6201.200
ee1102
cosc2803
math39512
omp9727
int2067/int5051
bsb151
mgt253
fc021
babs2202
mis2002s
phya21
18-213
cege0012
mdia1002
math38032
mech5125
07
cisc102
mgx3110
cs240
11175
fin3020s
eco3420
ictten622
comp9727
cpt111
de114102d
mgm320h5s
bafi1019
math21112
efim20036
mn-3503
fins5568
110.807
bcpm000028
info6030
bma0092
bcpm0054
math20212
ce335
cs365
cenv6141
ftec5580
math2010
ec3450
comm1170
ecmt1010
csci-ua.0480-003
econ12-200
ib3960
ectb60h3f
cs247—assignment
tk3163
ics3u
ib3j80
comp20008
comp9334
eppd1063
acct2343
cct109
isys1055/3412
math350-real
math2014
eec180
stat141b
econ2101
msinm014/msing014/msing014b
fit2004
comp643
bu1002
cm2030
联系我们
- QQ: 9951568
© 2021
www.rj363.com
软件定制开发网!