有穷自动机的构造与识别

#include <string.h>
#include <stdio.h>
#include <stdlib.h>
int main()
{
char p[30][30];                //存放文法
char q[30][30];
int line = 0;
int n;
int i, j;
int count =0;
int k,t = 0;
int flag = 0;
int l,m = 0;
char VN[30] = {'\0'};           //存放非终结符号
char VT[30] = {'\0'};           //存放终结符号
printf("请输入规则个数：");
scanf("%d",&n);
line = n;
for(i = 0; i < 30; i++)                  //给字符串数组p、q全部赋值为'\0'
for(j=0;j<30;j++)
{
p[i][j]='\0';
q[i][j]='\0';
}
printf("请输入文法:\n");
for(i = 0; i < line; i++)
{
scanf("%s",p[i]);
}      //把字符分为终结符号合非终结符号
l=0;
m=0;
for(i = 0;i < line; i++)
{
for(j = 0;j < 30&&(p[i][j] != '\0');j++)
{                   // 非终结符号放入数组VN中
if((p[i][j]<='z' && p[i][j]>='a')||(p[i][j]<='9' && p[i][j]>='0'))
{
flag = 0;
for(t=0; VN[t] != '\0';t++)
{
if(VN[t] == p[i][j])
{
flag = 1;
break;
}
}
if(flag == 0)
{
VN[l] = p[i][j];
l++;
}
}
// 终结符号放入数组VT中
if(p[i][j]<='Z' && p[i][j]>='A')
{
flag = 0;
for(t = 0; t<30&&(VT[t] != '\0'); t++)
{
if(VT[t] == p[i][j])
{
flag = 1;
break;
}
}
if(flag==0)
{
VT[m] = p[i][j];
m++;
}
}
}
} //把规则右部分分离放入数组q中
count = 0;
k =0;
for(i = 0;i < line;i++)
{
for(j = 4;j < 30 && (p[i][j] != '\0');j++)
{
if((p[i][j]<='z' && p[i][j]>='a')||(p[i][j]<='Z' && p[i][j]>='A')||(p[i][j]<='9'&& p[i][j]>='0'))
{
q[count][k] = p[i][j];
k++;
}
else
{
count++;
k =0;
}
}
count++;
k=0;
} //判断是确定的还是非确定的有穷状态自动机并进行前半部分打印
//判断依据：q数组中每一行字符串是否相同
flag = 0;
for(i = 0;i < count  i++)
{
for(j = i+1; j < count  j++)
{
if(strcmp(q[i],q[j]) == 0)
{
flag = 1;
break;
}
}
}
if(flag == 1)
{
printf("是非确定的有穷状态自动机，即NFA\n\n");
printf("构造的有穷状态自动机为：\n");
printf("NFA N =（K，∑，M，{S}，{Z}）\n");
}
else
{
printf("是确定的有穷状态自动机，即DFA\n\n\n");
printf("构造的有穷状态自动机为：\n");
printf("DFA D=（K，∑，M，{S}，{Z}）\n");
}
printf("其中，\nK = {S");
for(i = 0;i < 30 && (VT[i]!='\0');i++)
{
printf(",%c",VT[i]);
}
printf("}\n");
printf("∑={");
for(i = 0;i < 30 && (VN[i]!='\0');i++)
{
printf("%c",VN[i]);
}
printf("}\n"); //分离文法；
k = 0;
count = 0;
for(i = 0; i < line; i++)
{
j = 4;
while(p[i][j] != '\0')
{
if(k < 4)
{
q[count][k] = p[i][k];
k++;
}
else
{
if((p[i][j]<='z'&&p[i][j]>='a')||(p[i][j]<='Z'&&p[i][j]>='A')||(p[i][j]<='9' && p[i][j]>='0'))
{
q[count][k] = p[i][j];
k++;
j++;
}
if(p[i][j] == '|')
{
count++;
k = 0;
j++;
}
}
}
count++;
k=0;
}
printf("\n");
//打印M后半部分
printf("M:\n");
l =0;
while(VN[l] != '\0')
{
printf("M(S,%c)={",VN[l]);
for(i = 0; i < 30  i++)
{
for(j = 4; j<30 && (q[i][j] != '\0');j++)
{
if(VN[l]==q[i][j]&&(q[i][j+1] == '\0')&&(q[i][j-1] == '='))
printf("%c ",q[i][0]);
}
}
printf("}\t");
l++;
}
printf("\n");
l = 0;
k = 0;
while(VT[k] != '\0')
{
l =0;
while(VN[l] != '\0')
{
printf("M(%c,%c)={",VT[k],VN[l]);
for(i = 0; i < 30  i++)
{
for(j = 4; j<30 && (q[i][j] != '\0');j++)
{
if(VT[k] == q[i][j] && VN[l] == q[i][j+1])
printf("%c ",q[i][0]);
}
}
printf("}\t");
l++;
}
k++;
printf("\n");
}
system("pause");
}