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IP屬地：寧夏 上傳時(shí)間：2021-10-29 格式：DOC 頁(yè)數(shù)：19 大小：239KB 積分：11 舉報(bào) 版權(quán)申訴

1、一、 問(wèn)題的重述現(xiàn)今社會(huì)網(wǎng)購(gòu)已經(jīng)成為一種常見(jiàn)的消費(fèi)方式，隨著物流行業(yè)也漸漸興盛，送貨員需要以最快的及時(shí)將貨物送達(dá)，而且他們往往一個(gè)人送多個(gè)地方，這就迫切需要一種解決方案來(lái)預(yù)測(cè)出最短路線以期送貨耗時(shí)最少。我們需要解決的問(wèn)題是：1、 建立數(shù)學(xué)模型，分析送貨點(diǎn)以及路段長(zhǎng)（即圖論中的權(quán)）對(duì)于送貨時(shí)間的影響；2、 收集整理出現(xiàn)今對(duì)于tsp問(wèn)題的研究資料及成果。利用其中的方法解決本體送貨路線的設(shè)計(jì)；3、 寫(xiě)出一份報(bào)告，闡釋我們對(duì)該問(wèn)題的理解及部分解法。二、基本假設(shè)1、 假定送貨員行進(jìn)過(guò)程中的速度恒定2、 不考慮天氣、路段的好壞對(duì)送貨的影響3、 不考慮貨主因?qū)ω浳锏牟粷M而引起的對(duì)送貨員送貨時(shí)間的拖延4、 不

2、考慮送貨員在一個(gè)地方交送多個(gè)貨物所造成的時(shí)耗。為簡(jiǎn)化起見(jiàn)，交送多個(gè)貨物時(shí)只按一件貨物交接時(shí)間計(jì)時(shí)（即在每個(gè)地方只耽擱3分鐘）。三、符號(hào)假設(shè)符號(hào)含義d任意兩送貨點(diǎn)之間的距離x任意送貨點(diǎn)的橫坐標(biāo)y任意送貨點(diǎn)的縱坐標(biāo)u圖中的任意一點(diǎn)v圖中處u之外的點(diǎn)pu與v之間的權(quán)（路徑）a距離矩陣i送貨點(diǎn)j與i不相同的送貨點(diǎn)d1地點(diǎn)27和39之間的距離m所送貨物的質(zhì)量v所送貨物的體積s最優(yōu)線路的總路程v送貨員的速度=24km/ht送貨員沿最優(yōu)路線完成送貨任務(wù)所需總時(shí)間t每件貨物交接花費(fèi)時(shí)間=3min四、問(wèn)題的分析 本體中給出了一快遞公司的送貨圖要求送貨員將貨物送到指定地點(diǎn)并達(dá)最優(yōu)化的目的。在問(wèn)題一中，要求將一到三

3、十號(hào)貨物送到指定地點(diǎn)并返回。為解決此問(wèn)題，我們首先應(yīng)該考慮的問(wèn)題是送貨員能否將貨物一次性攜帶，其中包括貨物的重量和體積，因?yàn)檫@樣才能將時(shí)間縮到最短，而題目所給數(shù)據(jù)恰恰符合我們的期望。第二步，由于送貨員送貨速度恒定，故可將最快成路線問(wèn)題問(wèn)題轉(zhuǎn)化為最短線路問(wèn)題。第三步，就是將該實(shí)際問(wèn)題轉(zhuǎn)化為數(shù)學(xué)問(wèn)題，也就是問(wèn)題的簡(jiǎn)化與抽象：（1） 由于送貨點(diǎn)大小相對(duì)于路程相比，它們要小得多，故可抽象的縮成一個(gè)點(diǎn)。兩送貨點(diǎn)之間的公路可簡(jiǎn)化為直線段或曲線段。而直線段或曲線段的長(zhǎng)即為此段公路長(zhǎng)。于是送貨線路圖在數(shù)學(xué)上抽象為所謂“加權(quán)網(wǎng)絡(luò)圖”或稱“賦權(quán)圖”。送貨點(diǎn)成為圖的頂點(diǎn)。之間的公路直線或曲線稱為圖的邊。公路長(zhǎng)稱為此

4、邊的長(zhǎng)。先求出所有的權(quán)，其關(guān)系可表示為：權(quán)（即距離）的平方=橫坐標(biāo)之差的平方+縱坐標(biāo)之差的平方（2） 送貨問(wèn)題可歸結(jié)為圖上的優(yōu)化問(wèn)題：在給定加權(quán)網(wǎng)絡(luò)圖上尋找從給定點(diǎn)o出發(fā)，所有圖上的點(diǎn)至少一次，再回到給定點(diǎn)o，且使得總路程最少的閉曲線。（如圖所示為送貨員的線路）（注：圖中為實(shí)際坐標(biāo)的五分之一）五、模型的建立與求解問(wèn)題1 在問(wèn)題一中，要確定，送貨員可以一次性攜帶要送所有貨物不用回到o點(diǎn)去取貨物，求最快完成的線路方式可以簡(jiǎn)化為求從o點(diǎn)到另外22個(gè)地點(diǎn)的最短距離。(一）模型的建立在圖論模型中，最短距離的求解是非常經(jīng)典的問(wèn)題，在許多優(yōu)化問(wèn)題中有很多廣泛的應(yīng)用。問(wèn)題就是運(yùn)用圖論方法求解的一個(gè)經(jīng)典例子。所

5、謂的最短距離就是指在給定的賦權(quán)圖中指定應(yīng)一對(duì)定點(diǎn)u，v間眾多的路中，尋求一條邊權(quán)之和最小的路。這樣的路稱之為從u到v的在短距離。其求解公式為：d* d =(x1-x2)*(x1-x2)+(y1-y2)* (y1-y2) 為方便期間，我們定義圖中從u到v的路p（u，v）上所有邊權(quán)之和稱之為路p的權(quán)。從u到v的最短路的權(quán)稱之為u到v間的距離。利用這些距離我們可以得出一個(gè)矩陣。由于tsp問(wèn)題的復(fù)雜性，很難求出最優(yōu)解，我們可以采用lingo軟件求解上面我們已經(jīng)得出了一個(gè)距離矩陣a, d ij表示送貨點(diǎn)i和j之間的距離.設(shè)0-1矩陣x用來(lái)表示經(jīng)過(guò)的各送貨點(diǎn)之間的路段。設(shè)當(dāng)送貨點(diǎn)i不到送貨點(diǎn)j時(shí)xij為0

6、，相反，xij為1.考慮每個(gè)送貨點(diǎn)后只有一個(gè)送貨點(diǎn)，則1 i=1，2，3， （2）考慮每個(gè)送貨點(diǎn)前只有一個(gè)送貨點(diǎn)，類(lèi)比上式可得結(jié)果但僅以以上約束條件不能避免在一次送貨過(guò)程中產(chǎn)生多余一個(gè)連通回路。為此引入額外變量ui（i=1，2，3，n）附加以下約束條件，即ui-uj+n*xij<=n-1,1<i!=j<=n該約束條件的解釋：i與j不會(huì)構(gòu)成回路，有xij=1，xji=1，則ui-uj<=-1,uj-ui<=-1,從而有0<=-2，導(dǎo)致矛盾。i，j與k不會(huì)構(gòu)成回路，若構(gòu)成回路，有xij=1，xji=1,xki=1,則ui-uj<=-1, uj-uk<

7、=-1,uk-ui<=-1,從而有0<=-3，導(dǎo)致矛盾。其他情況以此類(lèi)推。于是可以得到模型。（二）模型的求解與驗(yàn)證由于送貨員在一點(diǎn)可能會(huì)送去不止一個(gè)貨物，故我們?cè)谀Ｐ椭锌梢砸暈樗拓泦T在該點(diǎn)只送去一個(gè)貨物，其質(zhì)量為在該點(diǎn)所送貨物質(zhì)量之和。下面我們通過(guò)對(duì)130號(hào)貨物的分析可知，從o點(diǎn)通過(guò)20個(gè)送貨點(diǎn)可以完成送貨過(guò)程。（此圖表中有一極其特殊的點(diǎn)-地點(diǎn)39，與其他點(diǎn)只有一條相連接的路線（只與27接路線），不符合tsp問(wèn)題的求解。故本應(yīng)該有的21個(gè)送貨點(diǎn)變?yōu)榱?0個(gè)，得到的總路程結(jié)果再加上二倍的d1(d1為27與39之間的距離）。因?yàn)橛行┴浳锼屯坏攸c(diǎn)，為方便編程我們將不同送貨地點(diǎn)依次編號(hào)

8、為0-20號(hào)。下面我們引入了130號(hào)貨物所要送達(dá)的地點(diǎn)及其對(duì)應(yīng)編號(hào)（表1）和各個(gè)貨物所到地點(diǎn)的位置坐標(biāo)（表2）：編號(hào)貨物號(hào)送達(dá)地點(diǎn)重量(公斤)體積(立方米)11132.500.031622180.500.035433311.180.024044261.560.035055212.150.030566141.720.010077171.380.010988231.400.042699320.700.04811010381.330.02191111451.100.02871212430.950.022813392.560.059514452.280.03011315422.850.01901643

9、1.700.078217320.250.04121418361.790.01841519272.450.04451620242.930.042021310.800.010822272.250.001823261.570.02101724342.800.01031825401.140.015526450.680.03821927491.350.014428320.520.002029232.910.04872030161.200.0429表1 各貨物號(hào)信息表 位置點(diǎn)x坐標(biāo)(米)y坐標(biāo)(米)191855002144556037270570437356705262099561008014357100

10、25228087160252591384526801011935305011785035451265854185137630520014134055325152125597516153657045171416573851888258075195855816520780835521127708560222200883523147659055247790933025443595252610860963527103851050028565976529258098653015659955表2 30個(gè)位置點(diǎn)的坐標(biāo)在此我們?cè)O(shè)任意兩點(diǎn)之間的直線距離為d，如果兩個(gè)點(diǎn)之間沒(méi)有直線距離，我們可是為這兩點(diǎn)之間為無(wú)窮

11、大（在矩陣中可視為5000000），如此可得到一個(gè)關(guān)于編號(hào)為130號(hào)送貨地點(diǎn)之間距離的矩陣（此矩陣為21x21的距離矩陣）。運(yùn)用microsoft visual c+軟件可求得任意兩點(diǎn)之間排成的矩陣如下表所示（已作微調(diào)：在圖表中，有一些比較特殊的點(diǎn)，雖然它們不能直接相連接，但是通過(guò)中間的一些點(diǎn)（在問(wèn)題一中不需要送去貨物 ），我們同樣可以到達(dá)目的地。對(duì)于這些特定的點(diǎn)，我們實(shí)施以上的方法，可能會(huì)造成較大的誤差。所以，我們得到具體最優(yōu)線路后，會(huì)對(duì)所得線路進(jìn)行優(yōu)化處理，減小誤差，使之得到的路線盡可能接近標(biāo)準(zhǔn)線路如地點(diǎn)32、35、38，45、50、49和13、19、24點(diǎn)進(jìn)行修正，可使得到的線路更加符合

12、問(wèn)題的要求）： 0,5000000,2182,5000000,1392,1796,5000000,5000000,5000000,5000000, 5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000, 5000000,5000000,5000000,5000000,0,3113,5000000,5000000,5000000,5000000,5000000,5000000, 5000000,5000000,5000000,5000000,5000000,5000000,5000000,5716,5000000,5000

13、000,5000000,5000000,2182,3113,0,5000000,5000000,5000000,5342,5000000,5000000,5000000, 5000000,5000000,5000000,5000000,5000000,2883,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,0,5000000,5000000,5000000,5000000,5000000, 5000000,5000000,5000000,5000000,5000000,3270,1067,5000000,5000

14、000, 5000000,5000000,5000000,1392,5000000,5000000,5000000,0,2191,5000000,5000000,5000000,5000000, 5000000,5000000,5000000,5000000,5000000,986,5000000,5000000,5000000, 5000000,5000000,1796,5000000,5000000,5000000,2191,0,3296,1823,5000000,5000000,5000000,5000000,5000000,5000000,2880,5000000,5000000,50

15、00000,5000000,5000000,5000000,5000000,5000000,5342,5000000,5000000,3296,0,2195,5000000,5000000, 5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000, 5000000,5000000,2607,5000000,5000000,5000000,5000000,5000000,1823,2195,0,1774,5000000, 5000000,5000000,5000000,5000000,5000000,5000000,5000

16、000,5000000, 5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,1774,0,1311, 5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000, 5000000,5000000,2097,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,1311,0,2523,5000000,5000000,5000000,5000000,

17、5000000,5000000,5000000, 5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000, 5000000,2523,0,5000000,2618,5000000,1537,5000000,5000000,5000000, 5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000, 5000000,5000000,5000000,0,50000

18、00,2351,3182,5000000,5000000,5000000, 3217,6170,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000, 5000000,5000000,2618,5000000,0,917,5000000,5000000,5000000,5000000, 5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000, 5000000,5000000,500000

19、0,2351,917,0,5000000,5000000,5000000,5000000, 5000000,1971,5000000,5000000,5000000,5000000,3270,5000000,2880,5000000,5000000,5000000, 5000000,1537,3182,5000000,5000000,0,5000000,5000000,5000000,5000000, 5000000,5000000,5000000,5000000,2883,1067,986,5000000,5000000,5000000,5000000, 5000000,5000000,50

20、00000,5000000,5000000,5000000,0,2846,3155,5000000, 5000000,5000000,5000000,5716,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,2846,0,5000000, 5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000, 5000000,500

21、0000,5000000,5000000,5000000,5000000,5000000,3155,5000000,0,1630,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000, 5000000,5000000,5000000,3217,5000000,5000000,5000000,5000000,5000000,1630,0,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,500000

22、0, 5000000,5000000,5000000,6170,5000000,1971,5000000,5000000,5000000, 5000000,5000000,0,5000000,5000000,5000000,5000000,5000000,5000000,5000000,2607,5000000,2097, 5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000, 5000000,5000000,5000000,0;記此距離矩陣記作為矩陣a （a為21x21的矩陣）。由于邏輯的復(fù)雜性和大量的計(jì)算工作，我們引

23、入了lingo程序，在問(wèn)題一中，根據(jù)tsp問(wèn)題的求解方法設(shè)計(jì)出lingo程序，將所求距離矩陣a帶入如下lingo程序可得如下(已縮減為如下列表)： x( a1, a5) 1.000000 1392.000 x( a2, a3) 1.000000 3113.000 x( a3, a1) 1.000000 2182.000 x( a4, a16) 1.000000 1067.000 x( a5, a6) 1.000000 2191.000 x( a6, a8) 1.000000 1823.000 x( a7, a21) 1.000000 2607.000 x( a8, a7) 1.000000 2

24、195.000 x( a9, a10) 1.000000 1311.000 x( a10, a11) 1.000000 2523.000 x( a11, a13) 1.000000 2618.000 x( a12, a15) 1.000000 3182.000 x( a13, a14) 1.000000 917.0000 x( a14, a20) 1.000000 1971.000 x( a15, a4) 1.000000 3270.000 x( a16, a17) 1.000000 2846.000 x( a17, a2) 1.000000 5716.000 x( a18, a19) 1.0

25、00000 1630.000 x( a19, a12) 1.000000 3217.000x( a20, a18) 1.000000 6610.000 x( a21, a9) 1.000000 2097.000有上述數(shù)據(jù)可排列成一組點(diǎn)的排列：a1a5a6a8a7a21a9a10a11a13a14a20a18a19a12a15a4a16a17a2a3a1由于lingo程序運(yùn)行的a對(duì)應(yīng)的數(shù)與c+中數(shù)組的數(shù)相差1，故可得出送貨地點(diǎn)的編號(hào)排列如下：045762089101213191718111431516120對(duì)照表1我們可以得出經(jīng)過(guò)的地點(diǎn)的流程圖如下：02621171416233238434249

26、34403627312413180由于在開(kāi)始我們?nèi)サ袅说攸c(diǎn)39后進(jìn)行解題，所以要補(bǔ)上地點(diǎn)39，得出送貨員的最優(yōu)路線依次為：02621171416233238434249344036273927312413180最優(yōu)線路的總路程s=1392.000+ 3113.000+ 2182.000+ 1067.000+ 2191.000+ 1823.000+ 2607.000+ 2195.000+1311.000+2523.000+2618.000+ 3182.000+ 917.0000+1971.000+ 270.000+ 2846.000+ 5716.000+ 1630.000+ 3217.000+

27、6610.000+ 2097.000+2xd=51478+2x1780=55038m送貨員在行程中所用時(shí)間=s/v=t3xt=2.293h（不包括交接貨物時(shí)間）送貨員在整個(gè)過(guò)程花費(fèi)時(shí)間（包括交接貨物時(shí)間）t=s/v+tx21=12035.7s=3.34325h 其線路如圖：355019181324394331423826273645403449322316141721六、模型的優(yōu)缺點(diǎn)優(yōu)點(diǎn)：在問(wèn)題中，我們分別是用了microsoft visual c+和lingo軟件，算出了圖中任意點(diǎn)之間的距離和所求的最短距離，簡(jiǎn)化了一些不必要的過(guò)程，使問(wèn)題得到了有效地解決。本文引用了tsp問(wèn)題和旅行售貨員問(wèn)題

28、，這兩個(gè)問(wèn)題都是經(jīng)典問(wèn)題，本文就是這兩問(wèn)題的推廣。本文引入了大量實(shí)際數(shù)據(jù)驗(yàn)證，從而保證了模型的準(zhǔn)確性，可以更好的適應(yīng)各種目標(biāo)。缺點(diǎn)： 問(wèn)題采用的解題方法得到的不是最優(yōu)解，而是近似最優(yōu)解，真正準(zhǔn)確的最優(yōu)解在實(shí)際中很難達(dá)到。 本題所建的模型還不夠成熟，在實(shí)際問(wèn)題中還需視情況進(jìn)行改進(jìn)。七、模型的改進(jìn)和推廣針對(duì)作業(yè)項(xiàng)目的最優(yōu)路線問(wèn)題，本文采用圖論的相關(guān)知識(shí)和兩個(gè)經(jīng)典問(wèn)題的解法有效的將問(wèn)題得到了轉(zhuǎn)化，建立了兩個(gè)模型，分別采用microsoft visual c+和lingo軟件，將問(wèn)題得到了解決，得到了最后的最有路徑。 在工程實(shí)際中，對(duì)于文中的路線的選擇問(wèn)題是很難實(shí)現(xiàn)達(dá)到最優(yōu)化的，較有效的一種方法還有模

29、糊分析法，multinomial logit模型分析法、基于rfm的分析法等。八、參考文獻(xiàn) 1姜啟源，謝金星，葉俊，數(shù)學(xué)模型（第三版），北京：高等教育出版社，2003年8月。2肖華勇，基于條件概率的定向促銷(xiāo)模型matlab和lingo的數(shù)學(xué)實(shí)驗(yàn)，西北工業(yè)大學(xué)出版社，2008年9月。3周曉陽(yáng)，數(shù)學(xué)實(shí)驗(yàn)與matlab,華中科技大學(xué)出版社，2001年4月。4張長(zhǎng)虹，數(shù)學(xué)建模簡(jiǎn)明教程，高等教育出版社，2008年5月。5王朝瑞 ，圖論，北京理工大學(xué)出版社，1981年。6劉乃實(shí)，言語(yǔ)幽默的圖論模型/當(dāng)代語(yǔ)言學(xué)研究文庫(kù)，上海交通大學(xué)出版社，2008年11月。附錄：下面運(yùn)用lingo軟件編譯的求解最短路徑的源

30、代碼（程序1）：model:sets:city/a1.a21/:u;!u(i)=cicy no;links(city,city):distance,!the distance of the matrix; x;!x(i,j)=1 if we use link i,j;endsetsdata: !distance matrix; !10000,the distance of impossible link;distance= 0,5000000,2182,5000000,1392,1796,5000000,5000000,5000000,5000000,5000000,5000000,50000

31、00,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,0,3113,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5716,5000000,5000000,5000000,5000000,2182,3113,0,5000000,5000000,5000000,5342,5000000,5000000,5000000,5000000,5000

32、000,5000000,5000000,5000000,2883,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,0,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,3270,1067,5000000,5000000,5000000,5000000,5000000,1392,5000000,5000000,5000000,0,2191,5000000,5000000,5000000,5000000,500

33、0000,5000000,5000000,5000000,5000000,986,5000000,5000000,5000000,5000000,5000000,1796,5000000,5000000,5000000,2191,0,3296,1823,5000000,5000000,5000000,5000000,5000000,5000000,2880,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5342,5000000,5000000,3296,0,2195,5000000,5000000,5000000

34、,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,2607,5000000,5000000,5000000,5000000,5000000,1823,2195,0,1774,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,1774,0,1311,5

35、000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,2097,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,1311,0,2523,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000

36、000,5000000,2523,0,5000000,2618,5000000,1537,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,0,5000000,2351,3182,5000000,5000000,5000000,3217,6170,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5

37、000000,5000000,5000000,2618,5000000,0,917,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,2351,917,0,5000000,5000000,5000000,5000000,5000000,1971,5000000,5000000,5000000,5000000,3270,5000000,2880,5000000,

38、5000000,5000000,5000000,1537,3182,5000000,5000000,0,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,2883,1067,986,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,0,2846,3155,5000000,5000000,5000000,5000000,5716,5000000,5000000,5000000,5000000,5000000,5

39、000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,2846,0,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,3155,5000000,0,1630,5000000,5000000,5000000,5000000,5000000,5000000,5000000,50

40、00000,5000000,5000000,5000000,5000000,5000000,3217,5000000,5000000,5000000,5000000,5000000,1630,0,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,6170,5000000,1971,5000000,5000000,5000000,5000000,5000000,0,5000000,5000000,5000000,5000000,500000

41、0,5000000,5000000,2607,5000000,2097,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,5000000,0;enddatan=size(city);!minimize total distance of the links;min=sum(links:distance*x);!the entrance;for(city(k):sum(city(i)|i#ne#k:x(i,k)=1;!it must be departed;sum(city(j)|j#n

42、e#k:x(k,j)=1;for(city(j)|j#gt#1 #and# j#ne#k:u(j)>=u(k)+x(k,j)-(n-2)*(1-x(k,j)+(n-3)*x(j,k););for(links:bin(x); !it makes the x's 0 or 1;for(city(k)|k#gt#1:u(k)<=n-1-(n-2)*x(1,k);u(k)>=1+(n-2)*x(k,1);end運(yùn)用microsoft visual c+軟件編譯的求距離矩陣a 的代碼（程序2）：#include<stdio.h>#include<math.h&

43、gt;#define inf 5e6;void main()int i,j;double c12121;int c2121;double zuobiaox21=11000,7630,8825,9395,10860,12770,13405,14165,14765,14835,13915,10915,14180,13265,12390,10385,7790,7280,8345,11575,15365; double zuobiaoy21=8250,5200,8075,10100,9635,8560,5325,7385,9055,10365,11610,14235,14215,14145,11415

44、,10500,9330,11065,12300,15160,7045;for(i=0;i<21;i+)for(j=0;j<21;j+)c1ij=sqrt(zuobiaoxi-zuobiaoxj)*(zuobiaoxi-zuobiaoxj)+(zuobiaoyi-zuobiaoyj)*(zuobiaoyi-zuobiaoyj); cij=(int)c1ij; c01=inf;c03=inf; c06=inf;c07=inf;c08=inf;c09=inf;c010=inf;c011=inf;c012=inf;c013=inf;c014=inf;c015=inf;c016=inf;c017=inf;c018=inf;c019=inf;c020=inf;c13=inf;c14=inf;c15=inf;c16=inf;c17=inf;c18=inf;c19=inf;c110=inf;c111=inf;c112=inf;c113=inf;c114=inf;c115=inf;c116=inf;c117=inf;c118=inf;c119=inf;c120=inf;c23=inf;c24=inf; c25=inf;c27=inf;c2