#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <time.h>

#define SUM 20            //總共的染色體數(shù)量
#define MAXloop 1200       //最大循環(huán)次數(shù)
#define error 0.01        //若兩次最優(yōu)值之差小于此數(shù)則認為結(jié)果沒有改變
#define crossp 0.7        //交叉概率
#define mp 0.04           //變異概率

struct gen                //定義染色體結(jié)構(gòu)
{
	int info;        
	float suitability;
};
struct gen gen_group[SUM];//定義一個含有20個染色體的組
struct gen gen_new[SUM];  

struct gen gen_result;    //記錄最優(yōu)的染色體
int result_unchange_time; //記錄在error前提下最優(yōu)值為改變的循環(huán)次數(shù)

struct log                //形成鏈表，記錄每次循環(huán)所產(chǎn)生的最優(yōu)的適應(yīng)度
{
	float suitability;
	struct log *next;
}llog,*head,*end;
int log_num;              //鏈表長度

void initiate();          
void evaluation(int flag);
void cross();
void selection();
int  record();
void mutation();
void showresult(int);

int   randsign(float p);
int   randbit(int i,int j);
int   randnum();
int   convertionD2B(float x);
float convertionB2D(int x);
int   createmask(int a);

void main()
{
	int i,flag;
	flag=0;
	initiate();
    evaluation( 0 );
	for( i = 0 ; i < MAXloop ; i++ )
	{
		cross();
		evaluation( 1 );
		selection();
		if( record() == 1 )
		{
			flag = 1;
			break;
		}
		mutation();
	}
	showresult( flag );
}

void initiate()
{
	int i , stime;	
	long ltime;
	ltime=time(NULL);
	stime=(unsigned)ltime/2;
	srand(stime);
	for( i = 0 ; i < SUM ; i++ )
	{
		gen_group[i].info = randnum();		 
	}
	gen_result.suitability=1000;
	result_unchange_time=0;
	head=end=(struct log *)malloc(sizeof(llog));
	if(head==NULL)
	{
		printf("\n內(nèi)存不夠！\n");
		exit(0);
	}
	end->next = NULL;
	log_num = 1;
}

void evaluation(int flag)
{
	int i,j;
	struct gen *genp;
	int gentinfo;
	float gentsuitability;
	float x;
	if( flag == 0 )
		genp = gen_group;
	else genp = gen_new;
	for(i = 0 ; i < SUM ; i++)//計算各染色體對應(yīng)的表達式值
	{
		x = convertionB2D( genp[i].info );
		genp[i].suitability = x*(x*(x*(x*(x*(x-10)-26)+344)+193)-1846)-1680;
	}
	for(i = 0 ; i < SUM - 1 ; i++)//按表達式的值進行排序，
	{
		for(j = i + 1 ; j < SUM ; j++)
		{
			if( genp[i].suitability > genp[j].suitability )
			{
				gentinfo = genp[i].info;
				genp[i].info = genp[j].info;
				genp[j].info = gentinfo;
				
				gentsuitability = genp[i].suitability;
				genp[i].suitability = genp[j].suitability;
				genp[j].suitability = gentsuitability;		
			}
		}
	}
}

void cross()
{
	int i , j , k ;
	int mask1 , mask2;
	int a[SUM];
	for(i = 0 ; i < SUM ; i++)  a[i] = 0;
	k = 0;
	for(i = 0 ; i < SUM ; i++)
	{
		if( a[i] == 0)
		{
			for( ; ; )//隨機找到一組未進行過交叉的染色體與a[i]交叉
			{
   				j = randbit(i + 1 , SUM - 1);
				if( a[j] == 0)	break;
			}
			if(randsign(crossp) == 1)
			{
				mask1 = createmask(randbit(0 , 14));
				mask2 = ~mask1;
				gen_new[k].info = (gen_group[i].info) & mask1 + (gen_group[j].info) & mask2;
				gen_new[k+1].info=(gen_group[i].info) & mask2 + (gen_group[j].info) & mask1;
				k = k + 2;
			}
			else 
			{
				gen_new[k].info=gen_group[i].info;
				gen_new[k+1].info=gen_group[j].info;
				k=k+2;
			}
			a[i] = a[j] = 1;
		}
	}
}

void selection()
{
	int i , j , k;
	j = 0;
	i = SUM/2-1;
	if(gen_group[i].suitability < gen_new[i].suitability)
	{
		for(j = 1 ; j < SUM / 2 ; j++)
		{
			if(gen_group[i+j].suitability > gen_new[i-j].suitability)
				break;
		}
	}
	else
		if(gen_group[i].suitability>gen_new[i].suitability)
		{
			for(j=-1;j>-SUM/2;j--)
			{
				if(gen_group[i+j].suitability<=gen_new[i-j].suitability)
					break;
			}
		}
	for(k=j;k<SUM/2+1;k++)
	{
		gen_group[i+k].info = gen_new[i-k].info;
		gen_group[i+k].suitability = gen_new[i-k].suitability;
	}	
}

int record()
{
	float x;	
	struct log *r;
	r=(struct log *)malloc(sizeof(llog));
	if(r==NULL)
	{
		printf("\n內(nèi)存不夠！\n");
		exit(0);
	}
	r->next = NULL;
	end->suitability = gen_group[0].suitability;
	end->next = r;
	end = r;
	log_num++;

	x = gen_result.suitability - gen_group[0].suitability;
	if(x < 0)x = -x;
	if(x < error)
	{
		result_unchange_time++;
		if(result_unchange_time >= 20)return 1;
	}
	else
	{
		gen_result.info = gen_group[0].info;
		gen_result.suitability = gen_group[0].suitability;
		result_unchange_time=0;
	}
	return 0;
}

void mutation()
{
	int i , j , m;
	float x;
	float gmp;
	int gentinfo;
	float gentsuitability;
	gmp = 1 - pow(1 - mp , 11);//在基因變異概率為mp時整條染色體的變異概率
	for(i = 0 ; i < SUM ; i++)
	{
		if(randsign(gmp) == 1)
		{
			j = randbit(0 , 14);
			m = 1 << j;
			gen_group[i].info = gen_group[i].info^m;
			x = convertionB2D(gen_group[i].info);
			gen_group[i].suitability = x*(x*(x*(x*(x*(x-10)-26)+344)+193)-1846)-1680;
		}
	}
	for(i = 0 ; i < SUM - 1 ; i++)
	{
		for(j = i + 1 ; j < SUM ; j++)
		{
			if(gen_group[i].suitability > gen_group[j].suitability)
			{
				gentinfo = gen_group[i].info;
				gen_group[i].info = gen_group[j].info;
				gen_group[j].info = gentinfo;
				
				gentsuitability = gen_group[i].suitability;
				gen_group[i].suitability = gen_group[j].suitability;
				gen_group[j].suitability = gentsuitability;
			}
		}
	}
}

void showresult(int flag)//顯示搜索結(jié)果并釋放內(nèi)存
{
	int i , j;
	struct log *logprint,*logfree;
	FILE *logf;
	if(flag == 0)
		printf("已到最大搜索次數(shù)，搜索失敗！");
	else 
	{
		printf("當(dāng)取值%f時表達式達到最小值為%f\n",convertionB2D(gen_result.info),gen_result.suitability);
		printf("收斂過程記錄于文件log.txt");
		if((logf = fopen("log.txt" , "w+")) == NULL)
		{
			printf("Cannot create/open file");
			exit(1);
		}
		logprint=head;
		for(i = 0 ; i < log_num ; i = i + 5)//對收斂過程進行顯示
		{
			for(j = 0 ; (j < 5) & ((i + j) < log_num-1) ; j++)
			{
				fprintf(logf , "%20f" , logprint->suitability);
				logprint=logprint->next;				
			}
			fprintf(logf,"\n\n");
		}
	}
	for(i = 0 ; i< log_num ; i++)//釋放內(nèi)存
	{
		logfree=head;
		head=head->next;
		free(logfree);
		fclose(logf);
	}
	getchar();
}

int randsign(float p)//按概率p返回1
{
	if(rand() > (p * 32768))
		return 0;
	else return 1;
}
int randbit(int i, int j)//產(chǎn)生在i與j之間的一個隨機數(shù)
{
	int a , l;
	l = j - i + 1;
	a = i + rand() * l / 32768;
	return a;
}
int randnum()
{
	int x;
	x = rand() / 2;
	return x;
}
float convertionB2D(int x)
{
	float y;
	y = x;
	y = (y - 8192) / 1000;
	return y;
	
}
int convertionD2B(float x)
{
	int g;
	g = (x * 1000) + 8192;
	return g;
}
int createmask(int a)
{
	int mask;
	mask=(1 << (a + 1)) - 1;
	return mask;
}