遗传算法解决TSP问题

  1 #include<iostream>
  2 #include<string>
  3 #include <cstdlib>
  4 #include<ctime>
  5 #include <cmath>
  6 using  std::string;
  7 
  8 struct Position
  9 {
 10     string EID;
 11     double x;
 12     double y;
 13     double z;
 14 
 15 };
 16 
 17 double TSP(int n, struct Position* position,int T,int population_num,int top_num,int * sort)
 18 {
 19     //T;//迭代次数；
 20     //population_num//种群个数
 21     //top_num 选取前边多少个
 22     int *population;//存放种群编码
 23     int *p;
 24     int getPerm(int *b, int n);
 25     population=(int*)malloc(sizeof(int)*population_num*n);//一维数组存放
 26     double *dis;
 27     dis=(double*)malloc(sizeof(double)*n*n);
 28     double *fitness;
 29     fitness=(double*)malloc(sizeof(double)*population_num);
 30 
 31     int *b;
 32     b=(int*)malloc(sizeof(int)*population_num);
 33 
 34 
 35     int *population1;
 36     population1=(int*)malloc(sizeof(int)*population_num*n);
 37 
 38     int* bb;
 39     bb=(int*)malloc(sizeof(int)*n);
 40 
 41     int* bbb;
 42     bbb=(int*)malloc(sizeof(int)*n);
 43 
 44     p=population;
 45     //产生50个初始种群
 46     srand(time(0));
 47     for(int i=0;i<population_num;i++)
 48     {
 49         getPerm(p,n);
 50         p=p+n;
 51     }
 52     //以上生成了50个初始种群，放在population
 53 
 54     //准备好距离矩阵
 55 
 56     int getDis(struct Position* position,double *dis,int n);
 57     getDis(position,dis,n);
 58     
 59     
 60     
 61 
 62 for(int t=0;t<T;t++)
 63 {
 64     //a计算适应度
 65     for(int i=0;i<population_num;i++)
 66     {
 67         fitness[i]=0;
 68         
 69         for(int j=i*n;j<n+i*n-1;j++)
 70         {
 71             int ii=population[j];
 72             int jj=population[j+1];
 73 
 74             fitness[i]+=dis[ii*n+jj];
 75         }
 76         int ii=population[i*n+n-1];
 77         int jj=population[i*n];
 78         fitness[i]+=dis[ii*n+jj];
 79     }
 80 
 81     
 82     //b对适应度进行排序
 83     int getIndex_double(double *a, int *b,int n);
 84 
 85 
 86     getIndex_double(fitness, b,population_num);
 87 
 88 
 89 
 90     //c选择比较好的个体，放入一个新的矩阵population1
 91     for(int i=0;i<population_num;i++)
 92     {
 93         memcpy(&population1[b[i]*n],&population[i*n],4*n);
 94     }
 95 
 96 
 97 
 98 
 99 
100     //选取比较好的top_num个体，进行交叉操作，放在从top_num+1开始的top_num/2个
101     for(int i=top_num;i<population_num;i=i+2)
102     {
103         memcpy(&population1[i*n],&population1[(i-top_num)*n],n);
104         memcpy(&population1[(i+1)*n],&population1[(i+1-top_num)*n],n);
105 
106         //选择交叉位置
107         int pos=rand()%n;
108         int tmp=population1[i*n+pos];
109         population1[i*n+pos]=population1[(i+1)*n+pos];
110         population1[(i+1)*n+pos]=tmp;
111 
112         //重新生成排列
113         int getIndex(int *a, int *b,int n);
114         getIndex(&population1[i*n],bb,n);
115         memcpy(&population1[i*n],bb,4*n);
116 
117         getIndex(&population1[(i+1)*n],bb,n);
118         memcpy(&population1[(i+1)*n],bb,4*n);
119 
120     }
121 
122 
123     //变异操作
124     //随机选取其中的以为进行变异
125     int pos_population=rand()%population_num;
126     int pos=rand()%n;
127     int value=rand()%n;
128     population1[pos_population*n+pos]=value;
129     int getIndex(int *a, int *b,int n);
130 
131     getIndex(&population1[pos_population*n],bbb,n);
132     memcpy(&population1[pos_population*n],bbb,4*n);
133 
134 
135 
136     memcpy(population,population1,n*population_num*4);
137 
138 
139 }
140     //保存在population1中，适应度在fitness中，排序在b
141     int i;
142     for(i=0;i<population_num;i++)
143     {
144         if(b[i]==0)
145             break;
146     }
147 
148     memcpy(sort,&population[i*n],n*4);
149     return fitness[i];
150      
151 }
152 
153 int getDis(struct Position* position,double *dis,int n)
154 {
155     for(int i=0;i<n;i++)
156     {
157         for(int j=0;j<n;j++)
158         {
159             dis[i*n+j]=sqrt((position[i].x-position[j].x)*(position[i].x-position[j].x)+
160                 (position[i].y-position[j].y)*(position[i].y-position[j].y)+
161                 (position[i].z-position[j].z)*(position[i].z-position[j].z));
162             
163         }
164     }
165     return 0;
166 }
167 
168 int getPerm(int *b, int n)
169 {
170     //得到一个随机排列
171     int getIndex(int *a, int *b,int n); 
172     int* a;
173     a=(int*)malloc(sizeof(int)*n);
174     for(int i=0;i<n;i++)
175         a[i]=rand()%n;
176     getIndex(a, b,n);
177 
178     return 0;
179 }
180 
181 //产生排列
182 int getIndex(int *a, int *b,int n)
183 {
184     for(int i=0;i<n;i++)
185     {
186         b[i]=0;
187     }
188     for(int i=0;i<n-1;i++)
189     {
190         for(int j=i+1;j<n;j++)
191         {
192             if(a[i]>=a[j])
193                 b[i]++;
194             else
195                 b[j]++;
196         }
197     }
198     return 0;
199 }
200 
201 int getIndex_double(double *a, int *b,int n)
202 {
203     for(int i=0;i<n;i++)
204     {
205         b[i]=0;
206     }
207     for(int i=0;i<n-1;i++)
208     {
209         for(int j=i+1;j<n;j++)
210         {
211             if((a[i]-a[j])>=0.0001)
212                 b[i]++;
213             else
214                 b[j]++;
215         }
216     }
217     return 0;
218 }

 1 #include"tsp.h";
 2 
 3 int main()
 4 {
 5     int n=6,i;
 6     struct Position* person;
 7     struct Position* p;
 8     int *sort;
 9     sort=(int*)malloc(sizeof(int)*n);
10     person=(struct Position*)malloc(sizeof(Position)*n);
11     p=person;
12 
13     p->x=0;p->y=0;p->z=0;p=p+1;
14     p->x=2;p->y=0;p->z=0;p=p+1;
15     p->x=3;p->y=1;p->z=0;p=p+1;
16     p->x=2;p->y=2;p->z=0;p=p+1;
17     p->x=0;p->y=2;p->z=0;p=p+1;
18     p->x=0;p->y=1;p->z=0;p=p+1;
19 
20     double dis=TSP(n, person,10,20,10,sort);
21 
22     printf("%.1f\n顺序为：",dis);
23 
24     for(int i=0;i<n;i++)
25     {
26         printf("%d ",sort[i]);
27     }
28     printf("\n");
29 
30     system("pause"); 
31     return 0;
32 }