#include<iostream.h>
#include<stdlib.h>
#include<time.h>
#include<math.h>
#define citynumber 5
#define Q 100
#define p 0.5
#define NM2 1000
#define A 1
#define B 5
int ccdi=-1;//全局变量,用在myrand()中
float myrand()//产生0-1随机数,100个,每调用一次,结果不同
{srand(time(0));
float my[100];
ccdi++;
if (ccdi==100)
ccdi=0;
for(int mi=0;mi<100;mi++)
{float fav=rand()%10000;
my[mi]=fav/10000;}
return my[ccdi];
}
double fpkij(double T[citynumber][citynumber],double n[citynumber][citynumber],int tabu[citynumber][citynumber],int k,int s,int i,int j )
//定义函数用于计算Pij
{
//double A=0.5,B=0.5;
double sumup,pkij,sumdown;
sumdown=0;
for(int aTi=0;aTi<citynumber;aTi++)
{
for(int aTj=0;aTj<citynumber;aTj++)
aT[aTi][aTj]=pow(T[aTi][aTj],A);
}
for(int bni=0;bni<citynumber;bni++)
{
for(int bnj=0;bnj<citynumber;bnj++)
bn[bni][bnj]=pow(n[bni][bnj],B);
}
for (int can=0;can<citynumber;can++)//判断,除掉已经走过的城市
{
if(can==tabu[k][ci])
{
aT[i][can]=0;bn[i][can]=0;
}
}
sumup=aT[i][j]*bn[i][j];
for(int tj=0;tj<citynumber;tj++)
sumdown=aT[i][tj]*bn[i][tj]+sumdown;
pkij=sumup/sumdown;
return pkij;
}
void main()
{ double city[citynumber][2]={{0,1},{0,2},{2,2},{2,4},{1,3}/*,{3,4},{4,7},{2,8},{3,9},{1,10},
{1,0},{2,1},{3,0},{4,9},{5,2},{6,2},{7,1},{8,6},{9,0},{10,3}*/}; /*城市坐标*/
double d[citynumber][citynumber]; //L[j][k]是城市j to k距离
for(int j=0;j<citynumber;j++)
{d[j][k]=sqrt((city[j][0]-city[k][0])*(city[j][0]-city[k][0])+(city[j][1]-city[k][1])*(city[j][1]-city[k][1]));
// cout<<d[j][k]<<" ";
}//cout<<"n";
} /*计算距离,从j城市到k城市*/
/* for (int cj=0;cj<10;cj++)
{float c=myrand();
cout<<c<<" "<<"n";}*///输出随机数
double n[citynumber][citynumber];
for(int ni=0;ni<citynumber;ni++)
{
for(int j=0;j<citynumber;j++)
}//cout<<"n";
} /*初始化visibility nij*/
double L[citynumber];
int shortest[citynumber];
double T[citynumber][citynumber];
for(int ti=0;ti<citynumber;ti++)
{
for (int j=0;j<citynumber;j++)
{
//cout<<T[ti][j]<<" ";
}//cout<<"n";
}/*初始化t*/
double changT[citynumber][citynumber];
//step2:
for(int NC=0;NC<NM2;NC++)
{ for(int cti=0;cti<citynumber;cti++)
{
for (int j=0;j<citynumber;j++)
{
changT[cti][j]=0;//cout<<changT[cti][j]<<" ";
}//cout<<"n";
} /*初始化changT*/
int tabu[citynumber][citynumber];//tabu[k][s]表示第k只蚂蚁,第s次循环所在的城市
for (int i=0;i<citynumber;i++)
tabu[tai][i]=0;
}
for (int tabui1=0;tabui1<citynumber;tabui1++)
tabu[tabui1][0]=tabui1;
/*for (tai=0;tai<citynumber;tai++)
{
for (int i=0;i<citynumber;i++)
cout<<tabu[tai][i]<<" ";cout<<"n";
}*/
//初始化tabu
for(int kk=0;kk<citynumber;kk++)
L[kk]=0;
//第三步开始
for(int s=0;s<citynumber-1;s++)
{
for(int k=0;k<citynumber;)
{
int ci,can;
float sumpk=0;
float pkij;
hq2: can++;
if (can==citynumber) can=0;
for (ci=0;ci<=s;ci++)
{if(can==tabu[k][ci]) goto hq2;}
pkij=fpkij(T,n,tabu,k,s,tabu[k][s],can);
sumpk=sumpk+pkij;
else goto hq2;
tabu[k][s+1]=can;
k++;
}
} //第三步完成
/*for (tai=0;tai<citynumber;tai++)
{
for (int i=0;i<citynumber;i++)
}*///输出一个循环后的tabu[][]
//第四步开始
for(int k4=0;k4<citynumber;k4++)
{
s44=s4+1;
if (s44==citynumber) s44=0;
L[k4]+=d[tabu[k4][s4]][tabu[k4][s44]];
}//cout<<L[k4]<<" ";
}//计算L[k]
float shortest1=0; int short2=0;//最短距离
for(ii=1;shorti<citber;shi++ )
{
shortest1=L[0];
if(L[shorti]<=shortest1)
{shortest1=L[shorti];short2=shorti;}
}
//cout<<L[sort2]<<"n";cout<<short2<<"n";
for(int shoi=0;shoi<ctynumber;shoi++)
{
shortest[shoi]=tabu[short2][shoi];
//cout<<shest[shoi]<<" ";
}
//cout<<"n";
for(int k41=0;k41<citynumber;k41++)
{for(int s41=0,ss=0;s41<citynumber;s41++)
{
ss=s41+1;
if (ss==citynumber) ss=0;
changT[tabu[k41][s41]][tabu[k41][ss]]+=Q/L[k41];
changT[tabu[k41][ss]][tabu[k41][s41]]=changT[tabu[k41][s41]][tabu[k41][ss]];
}
}
/* for(int cti4=0;cti4<citynumber;cti4++)
{
for (int j=0;j<citynumber;j++)
{cout<<changT[cti4][j]<<" ";}cout<<"n";
}*/
//第四步完
// 第五步开始
for(int i5=0;i5<citynumber;i5++)
{
for(int j5=0;j5<citynumber;j5++)
{
// cout<<T[i5][j5]<<" ";
}
//cout<<"n";
}
}
for(int shoi1=0;shoi1<citynumber;shoi1++)
{
cout<<city[shortest[shoi1]][0]<<" "<<city[shortest[shoi1]][1]<<" ";
}
}