欧美一级a免费放视频,欧美一级a免费放视频_丰满年轻岳欲乱中文字幕电影_欧美成人性一区二区三区_av不卡网站,99久久精品产品给合免费视频,色综合黑人无码另类字幕,特级免费黄片,看黃色录像片,色色资源站无码AV网址,暖暖 免费 日本 在线播放,欧美com

合肥生活安徽新聞合肥交通合肥房產(chǎn)生活服務(wù)合肥教育合肥招聘合肥旅游文化藝術(shù)合肥美食合肥地圖合肥社保合肥醫(yī)院企業(yè)服務(wù)合肥法律

代寫Implementation of Graph Algorithms

時(shí)間:2023-11-14  來(lái)源:合肥網(wǎng)hfw.cc  作者:hfw.cc 我要糾錯(cuò)


C++ Implementation of Graph Algorithms

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).

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 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).

加QQ:99515681 或郵箱:[email protected]   WX:codehelp

 

掃一掃在手機(jī)打開當(dāng)前頁(yè)
  • 上一篇:代寫CMPT 125,、c++設(shè)計(jì)編程代做
  • 下一篇:代寫CS170編程,、代做Java程序設(shè)計(jì)
  • 無(wú)相關(guān)信息
    合肥生活資訊

    合肥圖文信息
    出評(píng) 開團(tuán)工具
    出評(píng) 開團(tuán)工具
    挖掘機(jī)濾芯提升發(fā)動(dòng)機(jī)性能
    挖掘機(jī)濾芯提升發(fā)動(dòng)機(jī)性能
    戴納斯帝壁掛爐全國(guó)售后服務(wù)電話24小時(shí)官網(wǎng)400(全國(guó)服務(wù)熱線)
    戴納斯帝壁掛爐全國(guó)售后服務(wù)電話24小時(shí)官網(wǎng)
    菲斯曼壁掛爐全國(guó)統(tǒng)一400售后維修服務(wù)電話24小時(shí)服務(wù)熱線
    菲斯曼壁掛爐全國(guó)統(tǒng)一400售后維修服務(wù)電話2
    美的熱水器售后服務(wù)技術(shù)咨詢電話全國(guó)24小時(shí)客服熱線
    美的熱水器售后服務(wù)技術(shù)咨詢電話全國(guó)24小時(shí)
    海信羅馬假日洗衣機(jī)亮相AWE  復(fù)古美學(xué)與現(xiàn)代科技完美結(jié)合
    海信羅馬假日洗衣機(jī)亮相AWE 復(fù)古美學(xué)與現(xiàn)代
    合肥機(jī)場(chǎng)巴士4號(hào)線
    合肥機(jī)場(chǎng)巴士4號(hào)線
    合肥機(jī)場(chǎng)巴士3號(hào)線
    合肥機(jī)場(chǎng)巴士3號(hào)線
  • 上海廠房出租 短信驗(yàn)證碼 酒店vi設(shè)計(jì)