MATLAB implements ant colony algorithm (complete and can be used directly!!)

#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]<<"   ";
}

}