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要求从1开始,最终回到1,且路径有方向,要求所有路径都经过。
& X+ ?) m, g* S9 t+ t0 a6 ^+ [以下是lingo的代码* @2 j$ t7 a) N7 [% H2 k# C
sets:
1 ~; m6 e$ N9 o; o" y% fnodes/A,B,C,D,E,F,G,H,I,J,K,L/;
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150 1652 l5 O! }/ i6 S$ \* _ M
130 230 160; f6 ~# l# E+ S4 B( P
140 100( v# {8 b( ]: y, e; z) A
100 190
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170 144 128 218 174# E, `, i3 z {
200 122 109 185
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n = @size(nodes);
* x5 h' G6 `" _* m% Z1 `( D1 o$ Vmin = @sum(arcs: c * x);. s$ C* G/ d! e: c2 Z+ u4 ~
@for(nodes(i):
2 [+ |; l: R% y' w9 D@sum(arcs(i,j):x(i,j))=@sum(arcs(k,i):x(k,i))
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@for(arcs: @bnd(1,x,9)); ( N7 v# b+ ^8 O: t% d' c
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它只能算出路径的步数和路程,并不能得到线路。希望能用mma解决这个问题。 --------------------------------------------------------------------- 自己用mma写的程序对于四点还可以,但是扩展到12点实在繁琐: Y7 j% n! B$ I' ]! V- d/ V: H
P11 = {2, 5}; P12 = {150, 165}; P21 = {5, 6}; P22 = {230, 160}; P51 = {1, 6}; P52 = {165, 144}; P61 = {2, 5}; P62 = {170, 144};(*只考虑1、2、5、6四点,P11为第一点“可去往的点”,P12表述对应的路程*) open[q_] := Module[{i = 1, randomD, randomP, D, randomreal, p1, p2, p5, p6, c, u, Df = 80000, uf, path = {1}, pathf = {1}, pb},(*随机搜索*) For[r = 0, r < q, r++, {p1 = {}; p2 = {}; p5 = {}; p6 = {}; u = 0; D = 0; i = 1; randomP = P11; randomD = P12; While[ Length[p1] != 2 || Length[p2] != 2 || Length[p5] != 2 || Length[p6] != 2 || i != 1, randomreal = RandomInteger[{1, Length[randomP]}]; c = randomP[[randomreal]]; AppendTo[path, c];(*Print[path];*) Which[ i == 1 && Product[If[p1[] != c, 1, 0], {i, 1, Length[p1]}] == 1, {AppendTo[p1, c]}, i == 2 && Product[If[p2[] != c, 1, 0], {i, 1, Length[p2]}] == 1, {AppendTo[p2, c]}, i == 5 && Product[If[p5[] != c, 1, 0], {i, 1, Length[p5]}] == 1, {AppendTo[p5, c]}, i == 6 && Product[If[p6[] != c, 1, 0], {i, 1, Length[p6]}] == 1, {AppendTo[p6, c]}]; i = c; D = D + randomD[[randomreal]]; Which[i == 1, {randomP = P11, randomD = P12}, i == 2, {randomP = P21, randomD = P22}, i == 5, {randomP = P51, randomD = P52}, i == 6, {randomP = P61, randomD = P62}]; u = u + 1; If[u > 11, Break[]]; If[Df > D, {Df = D, uf = u, pathf = path, path = {1}}, path = {1}]; } ]; Print[Df, ",", uf, ",", pathf];
; s1 e9 e& }5 k1 T6 O- E' h; Kopen[1000]
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