/* Simulated Annealing(模拟退火算法) 求解旅行商问题(TSP) 网上给的数据是31个省会的坐标，蚁群算法得到的结果是：15378 我算的结果中，最好的一次是：15495 */ #include<iostream> #include<cstdio> #include<cstdlib> //#include<ctime> #include<cmath> #include<time.h> #define N 31 //城市个数 #define Tmax 8000 //初始温度 #define Tmin 1E-10 //终止温度 #define RATE 0.95 //温度衰减率 #define in_loop 13000 //内层循环次数 #define out_loop 2000 //外层循环次数 #define p_limit 10000 //概率选择次数using namespace std;//31个省会x和y的坐标 double x[N]={1304,3639,4177,3712,3488,3326,3238,4196,4312,4386,3007,2562,2788,2381,1332,3715,3918,4061,3780,3676,4029,4263,3429,3507,3394,3439,2935,3140,2545,2778,2370}; double y[N]={2312,1315,2244,1399,1535,1556,1229,1004,790,570,1970,1756,1491,1676,695,1678,2179,2370,2212,2578,2838,2931,1908,2367,2643,3201,3240,3550,2357,2826,2975};//两个城市之间的距离 double d[N][N];void init(){//初始化任意两个城市的距离for(int i=0;i<N;i++)for(int j=0;j<i;j++){d[i][j]=d[j][i]=sqrt((y[j]-y[i])*(y[j]-y[i])+(x[j]-x[i])*(x[j]-x[i]));}for(int i=0;i<N;i++)d[i][i]=0;srand((unsigned)time(0)); }//路径 class path{ public:path(){}~path(){}int city[N]; //依次经过的城市编号double dis; //总的距离void totalLen(){dis=0;for(int i=0;i<N-1;i++)dis+=d[city[i]][city[i+1]];dis+=d[city[0]][city[N-1]];} };//最优解 path bestpath;//产生新路径 path newpath(path prepath){path newp = prepath;int x,y,t;do{x=rand()%N;y=rand()%N;}while(x==y);t=newp.city[x];newp.city[x]=newp.city[y];newp.city[y]=t;newp.totalLen();return newp; }//Annealing void Anne() {init();//随机选一个初始解for(int i=0;i<N;i++)bestpath.city[i]=i;bestpath.totalLen();int out_t=0,p_t=0;double T=Tmax,delta,prob, rnd;path np,cp; //新路径，当前路径cp=bestpath;while(out_t<out_loop&&T>Tmin){for(int i=0;i<in_loop;i++){np=newpath(cp);if(np.dis<cp.dis){cp=np;out_t=0;p_t=0;}else{delta=np.dis-cp.dis;prob=exp(-delta/T);rnd=rand()%10000/10000.0;if(prob>rnd)cp=np;p_t++;}if(p_t>p_limit){out_t++;break;}}if(cp.dis<bestpath.dis)bestpath=cp;T*=RATE;printf("dis= %f\n",bestpath.dis);} }int main() {Anne();return 0; }
