Mathematica 能否完全替代Lingo？

入戏太深丶

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。
% w2 h- P9 i) W以下是lingo的代码
sets:
nodes/A,B,C,D,E,F,G,H,I,J,K,L/;
1 j6 W8 S; M+ I& e% Karcs(nodes,nodes)/
# b3 h7 F7 i$ I# m0 p  e  ]A,B A,E
7 t  A; p& j: v: a1 n5 yB,C B,E B,F
C,B C,D
; g5 |. ?8 i6 ]% x5 x6 @- kD,C D,H
E,A E,F
F,B F,E F,G F,I F,J
, K- ~. I+ ]9 t4 Q+ CG,C G,F G,H G,K
H,D H,K H,L
I,E I,J
& [2 H; ?" l, c) P: KJ,F J,G
K,G K,J
' Q7 G% ?' ^) j7 P; r$ y8 dL,K
/: c, x;
endsets
data:
c=
150 165
7 g8 ~' Y( {+ e. O( o- x130 230 160
140 100
100 190
165 144
# }+ D  B5 A& H$ a170 144 128 218 174
200 122 109 185
0 @5 p9 o, @0 T0 U# [+ g6 b. k8 n180 141 190
194 148
0 O" w" W! G& d- W4 z174 233
, E, H% k* X) V5 W2 ?/ P185 135
110;
4 @0 l$ i0 J) b! ~enddata
$ M( B2 }5 \- J- N  M. Yn = @size(nodes);
+ c+ g; v/ u& Qmin = @sum(arcs: c * x);
% T6 _1 x, ]3 l. ]& V- [, q% o. k7 k@for(nodes(i):
@sum(arcs(i,j):x(i,j))=@sum(arcs(k,i):x(k,i))
);
( x4 ]4 @4 ~0 V2 q* T% `6 D0 ]; M5 a@for(arcs: @bnd(1,x,9));

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。

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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：

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];

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open[1000]

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