模拟退火算法求解TSP问题
1.问题
我方有一个基地,经度和纬度为(70, 40).假设我方飞机速度为1000km/h.我方派一架飞机从基地出发,侦查完所有目标,再返回原来的基地.再每一目标点的侦查时间不计,求该飞机所花费的最短时间.
已知两点经纬度为(x1, y1), (x2, y2), 则两点再地球表面的距离为
d = Rarccos(cos(x1 - x2)*cos(y1)*cos(y2) + sin(y1)*sin(y2))
2.数据
53.7121 15.3046 51.1758 0.0322 46.3253 28.2753 30.3313 6.9348 56.5432 21.4188 10.8198 16.2529 22.7891 23.1045 10.1584 12.4819 20.1050 15.4562 1.9451 0.2057 26.4951 22.1221 31.4847 8.9640 26.2418 18.1760 44.0356 13.5401 28.9836 25.9879 38.4722 20.1731 28.2694 29.0011 32.1910 5.8699 36.4863 29.7284 0.9718 28.1477 8.9586 24.6635 16.5618 23.6143 10.5597 15.1178 50.2111 10.2944 8.1519 9.5325 22.1075 18.5569 0.1215 18.8726 48.2077 16.8889 31.9499 17.6309 0.7732 0.4656 47.4134 23.7783 41.8671 3.5667 43.5474 3.9061 53.3524 26.7256 30.8165 13.4595 27.7133 5.0706 23.9222 7.6306 51.9612 22.8511 12.7938 15.7307 4.9568 8.3669 21.5051 24.0909 15.2548 27.2111 6.2070 5.1442 49.2430 16.7044 17.1168 20.0354 34.1688 22.7571 9.4402 3.9200 11.5812 14.5677 52.1181 0.4088 9.5559 11.4219 24.4509 6.5634 26.7213 28.5667 37.5848 16.8474 35.6619 9.9333 24.4654 3.1644 0.7775 6.9576 14.4703 13.6368 19.8660 15.1224 3.1616 4.2428 18.5245 14.3598 58.6849 27.1485 39.5168 16.9371 56.5089 13.7090 52.5211 15.7957 38.4300 8.4648 51.8181 23.0159 8.9983 23.6440 50.1156 23.7816 13.7909 1.9510 34.0574 23.3960 23.0624 8.4319 19.9857 5.7902 40.8801 14.2978 58.8289 14.5229 18.6635 6.7436 52.8423 27.2880 39.9494 29.5114 47.5099 24.0664 10.1121 27.2662 28.7812 27.6659 8.0831 27.6705 9.1556 14.1304 53.7989 0.2199 33.6490 0.3980 1.3496 16.8359 49.9816 6.0828 19.3635 17.6622 36.9545 23.0265 15.7320 19.5697 11.5118 17.3884 44.0398 16.2635 39.7139 28.4203 6.9909 23.1804 38.3392 19.9950 24.6543 19.6057 36.9980 24.3992 4.1591 3.1853 40.1400 20.3030 23.9876 9.4030 41.1084 27.7149
3.代码
%clc;clear; R = 6730; sj0 = load('sj.txt'); % load initial data the 0 in 'sj0' may mean initial x = sj0(:, [1:2:8]);x = x(:); y = sj0(:, [2:2:8]);y = y(:); % x = x(:) and y = y(:) will transfer matrix into a n*1 matrix without loss of data sj = [x, y]; d1 = [70, 40]; % d1 is both the beginning and end sj = [d1; sj; d1]; sj = sj*pi/180; % deg2rad() d = zeros(102); for i = 1:101 for j = i+1:102 d(i, j) = R*acos(cos(sj(i, 1) - sj(j, 1))*cos(sj(i, 2))*cos(sj(j, 2)) + sin(sj(i, 2))*sin(sj(j, 2))); d(i, j) = R*acos(cos(sj(i, 1) - sj(j, 1))*cos(sj(i, 2))*cos(sj(j, 2)) + sin(sj(i, 2))*sin(sj(j, 2))); end end d = d + d'; % magical path = []; long = inf; rand('state', sum(clock)); for j = 1:100 path0 = [1, 1 + randperm(100), 102]; temp = 0; for i = 1:101 temp = temp + d(path0(i), path0(i+1)); end if temp < long path = path0; long = temp; end end e = 1e-30; L = 20000; alpha = 0.999; T = 1; for k = 1:L c = 2 + floor(100*rand(1, 2)); c = sort(c); c1 = c(1); c2 = c(2); df = d(path(c1-1), path(c2)) + d(path(c1), path(c2 + 1)) - d(path(c1 - 1), path(c1)) - d(path(c2), path(c2 + 1)); if df < 0 | exp(-df/T) >= rand path = [path(1:c1 - 1), path(c2: -1: c1), path(c2+1: 102)]; long = long + df; end T = T*alpha; if T < e break; end end path,long xx = sj(path, 1); yy = sj(path, 2); plot(xx, yy, '-*');