数学建模旅游问题C4(6)

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e = 5;

Maxtime = 200; n = length(D); m = n;

Ta = Tao0*ones(n,n) - Tao0*eye(n); TM = []; LM = 100000; ant = zeros(m,n+1); ata = zeros(m,n); for i = 1:m for j = i+1:n

ata(i,j) = 1/D(i,j); ata(j,i) = ata(i,j); end end

for time = 1:Maxtime for k = 1:m

city(k,:) = 1:n; end s = 1; for i = 1:m

ant(i,s) = unidrnd(n); ant(i,n+1) = ant(i,s); city(i,ant(i,s)) = 0; end while s < n

p = zeros(m,n); for i = 1:m A(i) = 0; for j = 1:n

if city(i,j) ~= 0

A(i) = A(i) + (Ta(ant(i,s),j)^alpha)*(ata(ant(i,s),j)^beta); end end

for j = 1:n

if city(i,j) ~= 0

p(i,j) = (Ta(ant(i,s),j)^alpha)*(ata(ant(i,s),j)^beta)/A(i); end end

pp = unifrnd(0,1); j = 1; while pp>0

pp = pp - p(i,j); j = j + 1;

最佳旅游路线设计

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end j = j - 1; ant(i,s+1) = j; for k = 1:n

if city(i,k) == j; city(i,k) = 0; break; end end end s = s + 1; end

L = compu(ant,D); [Lmin,t] = min(L); if Lmin < LM LM = Lmin; TM = ant(t,:); end

deltaTa = zeros(m,n); for i = 1:m for j = 1:n-1

deltaTa(ant(i,j),ant(i,j+1)) = deltaTa(ant(i,j),ant(i,j+1)) + Q/L(i); end end

for j = 1:n-1

deltaTa(TM(j),TM(j+1)) = deltaTa(TM(j),TM(j+1)) + Q/LM; end for i = 1:m for j = 1:n

Ta(i,j) = rho*Ta(i,j) + deltaTa(i,j); end end

trace(time) = LM; end figure(1); plot(trace); toc;

function D = Distance(C) n = length(C); D = zeros(n,n); for i = 1:n for j = i+1:n

D(i,j) = norm(C(:,i) - C(:,j));

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D(j,i) = D(i,j); end end

function L = compu(ant,D); for i = 1:length(D) L(i) = 0;

for j = 1:length(ant)-1

L(i) = L(i) + D(ant(i,j),ant(i,j+1)); end end

程序3：问题一的MATLAB遗传算法程序 function [ y , road ] = GATSP1 % 程序说明：

% 本程序为遗传算法求解旅行商问题 %

% 变量说明： % 输入：

% D：城市位置矩阵n*2。 % 输出：

% y：近似最短路径的总长度； % road：近似最短路径。 % tic D=[

0.00 6.73 15.97

28.96 32.82 29.87 33.02 35.14 36.21 37.55 18.03 22.27

9.90 12.02

36.58

11.03 37.49

12.97 39.14 6.98 14.58 12.20 10.15

21.39

22.23 24.29

29.25 22.25

29.85 29.27

18.91

13.71

21.31

26.06

35.98 34.47

7.00 14.60 17.75

35.72 29.65 27.53 26.46 38.15 25.05 22.87

37.37

18.22

25.40

32.42

34.54 35.61

31.30

12.27 13.34

19.87 20.94

16.10 15.03

27.52 28.59

29.18 28.11

5.13 3.01 1.94 0.00 20.48 28.40 25.65 31.46

30.05

17.47 13.34

18.54 15.28

20.48

0.00 5.20 10.38

22.81

34.24 31.54

34.32

5.20 0.00 7.60 20.03

15.25

15.35 12.27

15.28

22.88

16.97

30.53

6.73 0.00 9.24 3.17 5.29 6.36 8.30 12.18 15.97

9.24 0.00 6.07 8.19 9.26 11.20

9.90 3.17 6.07 0.00 2.12 3.19 5.13 15.35 12.02 11.03 12.97 18.91 13.71

5.29 8.19 2.12 0.00 1.07 3.01 17.47 6.36 9.26 3.19 1.07 0.00 1.94 18.54 8.30 11.20 12.18

12.20

6.98 7.00 10.15

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21.31 14.58 14.60 17.75 19.87 20.94

22.88

10.38

7.65 14.67 15.27 23.86 23.94 26.06 21.39 24.29 18.22 16.10 15.03 16.97

22.81

0.00 20.08 22.35 21.75 11.43 13.08 28.96 22.23 22.25 25.40 27.52 28.59

30.53

18.03

20.08 0.00 7.02 7.62 21.07 16.29 35.98 29.25 29.27 32.42 34.54 35.61

37.55

25.05

22.35 7.02 0.00 0.60 14.05 9.27 36.58 29.85 29.87 33.02 35.14 36.21

38.15

25.65

21.75 7.62 0.60 0.00 13.45 8.67 37.49 32.82 35.72 29.65 27.53 26.46 28.40 34.24 11.43 21.07 14.05

13.45

0.00 4.78

39.14

34.47

37.37

31.30

29.18

28.11

30.05

34.32

13.08

16.29

9.27 8.67 4.78 0.00

];

NIND = 100; %种群数量； MAXGEN = 500; %最大遗传代数； pc = 0.8; %交叉概率； pm = 0.1; %变异概率；

N = length(D(:,1)); %求城市数量 %随机生成初始种群 for i = 1:NIND B(i,:) = manu(N);

F(i) = Distance(B(i,:),N,D); end SelCh = B; F = F';

FitnV = (10000./F); %计算适应度

for gen = 1:MAXGEN

for t=1:floor(NIND/30) %寻找每代的一部分最优群种以便遗传给下一代 for i = 1:NIND if F(i) == min(F)

bestSelCh(t,:) = SelCh(i,:); bestFit(t) = F(i); break; end end end

SelCh = select ('rws', SelCh, FitnV); %选择

for i = 1:0.1*NIND %生成一部分子代 Chrom(i,:) = manu(N);

C(i) = Distance(Chrom(i,:),N,D); G(i) = 10000/C(i); 最佳旅游路线设计

7.60 0.00 12.43

20.03

12.43

15.25

7.65

22.27

14.67

22.87

15.27

31.46 23.86 31.54

23.94

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end

SelCh = reins(SelCh,Chrom,1,1,F,G'); %插入 SelCh = Crossover (SelCh,N,NIND,pc); %交叉 SelCh = Muta (SelCh,N,NIND,pm); %变异 for i = 1:NIND

F(i) = Distance(SelCh(i,:),N,D); end

for t=1:floor(NIND/30) %遗传的最优个体替换适应度最差的新个体 for i = 1:NIND if F(i) == max(F)

SelCh(i,:) = bestSelCh(t,:); F(i) = bestFit(t); break; end end end

FitnV = (10000./F); %计算适应度 trace (gen, 1) = min(F); %寻找每一代的最优值 for i = 1:NIND

if F(i) == trace (gen, 1) key = SelCh(i,:); break; end end

trace (gen, 2) = mean(F); %寻找每一代的平均值 if gen == 1

y = trace (gen, 1); road = key; else

if y > trace (gen, 1) y = trace (gen, 1);

road = key; %寻找当前最优路径 end end end toc

%作图，观察遗传代数与路径长度的关系 figure(2); hold on;

plot(trace(:,1)); plot(trace(:,2),'r'); title('性能追踪'); xlabel('进化代数');

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