題目：利用遺傳算法求解區間[0,31]上的二次函數y=x*x的最大值。
//題目: 利用遺傳算法求解區間[0, 31]上的二次函數y = x*x的最大值
#include
#include
#include
#include
using namespace std;
int N = 4;//種羣規模
int T = 100;//最大換代數
double Pc = 0.4;//交叉率
double Pm = 0.01;//變異率
int option = 0;
int t;//代數計數器
bool flag = true;//標記是否是第一次計算結果
struct solution{
int num; //十進制數
char b[5]; //5位二進制數編碼染色體
int f; //適應度
double p; //選擇概率
double q; //積累概率
}*s, *next;
void showCondition(){
cout << "種羣規模: " << N << endl;
cout << "最大換代數: " << T << endl;
cout << "交叉率: " << Pc << endl;
cout << "變異率: " << Pm << endl << endl;
if(!flag)
system("pause");
}
//交換字符
void swap(char *a, char *b){
char c = *a;
*a = *b;
*b = c;
}
//十進制轉換爲二進制
void DtoB(int n, char str[]){
int tmp, i = 0;
memset(str, '0', 5);
while(n){
tmp = n % 2;
str[i] = (char)tmp;
n /= 2;
i++;
}
//翻轉
swap(&str[0], &str[4]);
swap(&str[1], &str[3]);
}
//生成初始種羣
void init(){
t = 1;
srand((unsigned)time(0));
if(flag){
s = new struct solution[N];
next = new struct solution[N];
flag = false;
}
else{
delete [] s;
delete [] next;
s = new struct solution[N];
next = new struct solution[N];
}
for(int i = 0; i < N; i++){
s[i].num = rand() % 32;
DtoB(s[i].num, s[i].b);
s[i].f = s[i].num * s[i].num;
}
int sumf = 0;//適應度總和
for(int i = 0; i < N; i++)
sumf += s[i].f;
double sumq = 0.0;//積累概率總和
for(int i = 0; i < N; i++){
s[i].p = (double)s[i].f / sumf;
sumq += s[i].p;
s[i].q = sumq;
}
}
//產生0~1之間的隨機數
double randf()
{
return (double)(rand() % 1001) * 0.001;
}
//選擇-複製
void SelectNCopy(){
double r;
struct solution tmp[N];
for(int i = 0; i < N; i++){
r = randf();
if(r < s[0].q)
tmp[i] = s[0];
else{
int j;
for(j = 1; j < N; j++){
if(r > s[j-1].q && r < s[j].q)
break;
}
tmp[i] = s[j];
}
}
for(int i = 0; i < N; i++)
next[i] = tmp[i];
}
//交叉
void mate(){
int a = 0, b = 0;
while(a == b){
srand((unsigned)time(0));
a = rand() % N;
b = rand() % N;
}
int cb = rand() % 5;//交換的位數
char ctmp;
for(int i = 0; i < cb; i++) {
// swap(&s[a].b[4 - i], &next[b].num[4 - i]);
ctmp = s[a].b[4 - i];
next[a].b[4 - i] = next[b].b[4 - i];
next[b].b[4 - i] = ctmp;
}
}
//變異
void variate(){
int bits = int(5 * Pm);
if(bits < 1)
return ;
else{
int tmp;
srand((unsigned)time(0));
for(int i = 0; i < N; i++){
tmp = rand() % 5;
if(next[i].b[tmp] == '0')
next[i].b[tmp] = '1';
else
next[i].b[tmp] = '0';
}
}
return ;
}
//返回適應度最大的解
struct solution maxf(struct solution ss[]){
struct solution temp = ss[0];
for(int i = 1; i < N; i++) if(ss[i].f > temp.f)
temp = ss[i];
return temp;
}
//進化，適者生存
void evolve(){
struct solution cm, nm;// cm：當前代適應度最大值，nm下一代適應度最大值
cm = maxf(s) ;
nm = maxf(next);
if(nm.f > cm.f)
for(int i = 0; i < N; i++)
s[i] = next[i];
}
//用遺傳算法求解答案
void solve(){
int mt;//交叉次數
while(t SelectNCopy();
mt = (int)(5 * Pc) / 2;
for(int i = 0; i < mt; i++)
mate();
variate();
evolve();
t++;
}
struct solution max = s[0];
for(int i = 0; i < N; i++) if(s[i].f > max.f)
max = s[i];
printf("求得的解是：%d\n", max.num);
return ;
}
// 計算答案
void calculate(){
init();
solve();
}
void menu(){
cout << "利用遺傳算法求解區間[0, 31]上的二次函數y = x*x的最大值\n" << endl;
cout << "初始狀態：" << endl;
showCondition();
cout << endl;
while(true){
cout << "1>>>計算結果" << endl;
cout << "2>>>設置種羣規模" << endl;
cout << "3>>>設置最大換代數" << endl;
cout << "4>>>設置交叉率" << endl;
cout << "5>>>設置變異率" << endl;
cout << "6>>>顯示各種參數" << endl;
cout << "0>>>退出程序\n" << endl;
cin >> option;
switch(option){
case 1:calculate();break;
case 2:
cout << "請輸入新的種羣規模值： ";
cin >> N;break;
case 3:
cout << "請輸入新的最大換代數： ";
cin >> T;break;
case 4:
cout << "請輸入新的交叉率： ";
cin >> Pc;break;
case 5:
cout << "請輸入新的變異率： ";
cin >> Pm;break;
case 6:showCondition();break;
case 0:exit(0);break;
default:cout << "選擇錯誤！請重新選擇！\n" << endl;break;
}
}
}
int main(){
menu();
return 0;
}