1118有限自动机的识别与构造

  1 #include<stdio.h>
  2 #define Max 100
  3 int flag=0;
  4 struct data
  5 {
  6     int a;
  7     char r;
  8     int b;
  9 };
 10 
 11 typedef struct data Data;
 12 void First(int k,int Mark1[],int arry[]);
 13 void Second(int k,int Mark2[],int arry[]);
 14 void Fird(int k,int Mark3[],int arry[]);
 15 void Fouth(int k,int Mark4[],int arry[]);
 16 void Fifth(int k,int Mark5[],int arry[]);
 17 
 18 main()
 19 {
 20     Data fa[Max];
 21     char ch;
 22     char string[Max];
 23     int Mark1[Max];
 24     int Mark2[Max];
 25     int Mark3[Max];
 26     int Mark4[Max];
 27     int Mark5[Max];
 28     int  arry[Max];
 29     
 30     //int flag=0;
 31     int i=-1,j=0,k=0,m=0,n=0;
 32     int count1=0;
 33     int count2=0;
 34     int count3=0;
 35     int count4=0;
 36     int count5=0;
 37 
 38     printf("请输入正规式：");
 39     do{
 40         i++;
 41         scanf("%c",&ch);
 42         string[i]=ch;
 43     }while(ch!='#');
 44     string[i]='\0';
 45     for(j=0;j<i;j++)
 46     {
 47         if(string[j]=='|')
 48         {
 49             Mark1[count1]=j;
 50             count1++;
 51         }
 52         else if(string[j]=='.')
 53         {
 54             Mark2[count2]=j;
 55             count2++;
 56         }
 57         else if(string[j]=='*')
 58         {
 59             Mark3[count3]=j;
 60             count3++;
 61         }
 62         else if(string[j]=='(')
 63         {
 64             Mark4[count4]=j;
 65             count4++;
 66         }
 67         else if(string[j]==')')
 68         {
 69             Mark5[count5]=j;
 70             count5++;
 71         }
 72 
 73         else
 74             continue;
 75     }
 76 
 77     while(m!=i)
 78     {
 79         for(k=0;k<count1;k++)
 80             First(k,Mark1,arry);
 81         for(k=0;k<count2;k++)
 82             Second(k,Mark2,arry);
 83         for(k=0;k<count3;k++)
 84             Fird(k,Mark3,arry);
 85         for(k=0;k<count4;k++)
 86             Fouth(k,Mark4,arry);
 87         for(k=0;k<count5;k++)
 88             Fifth(k,Mark5,arry);
 89         m++;
 90     }
 91 
 92 
 93     printf("f(%d,%c)=%d\n",fa[i].a,fa[i].b,fa[i].r);
 94 
 95 
 96 
 97     //printf("%s\n",string);
 98     //for(k=0;k<count1;k++)
 99     //printf("%d\n",Mark1[k]);
100 }
101 void First(int k,int Mark1[],int arry[])
102 {
103 
104 
105 }
106 void Second(int k,int Mark2[],int arry[])
107 {
108 
109 
110 }
111 void Fird(int k,int Mark3[],int arry[])
112 {
113 
114 
115 }
116 void Fouth(int k,int Mark4[],int arry[])
117 {
118 
119 
120 }
121 void Fifth(int k,int Mark5[],int arry[])
122 {
123 
124 }

  1 #include <string.h> 
  2 #include <stdio.h> 
  3 #include <stdlib.h>
  4 int main() 
  5 {      
  6     char p[30][30];                //存放文法    
  7     char q[30][30];                    
  8     int line = 0;                         
  9     int n;      
 10     int i, j;       
 11     int count =0;   
 12     int k,t = 0;      
 13     int flag = 0;      
 14     int l,m = 0;     
 15     char VN[30] = {'\0'};           //存放非终结符号    
 16     char VT[30] = {'\0'};           //存放终结符号             
 17     printf("请输入规则个数：");       
 18     scanf("%d",&n);       
 19     line = n;            
 20     for(i = 0; i < 30; i++)                  //给字符串数组p、q全部赋值为'\0'  
 21    for(j=0;j<30;j++)     
 22    {    
 23        p[i][j]='\0';                  
 24        q[i][j]='\0';          
 25    }       
 26    printf("请输入文法:\n");        
 27    for(i = 0; i < line; i++)       
 28    {      
 29        scanf("%s",p[i]);       
 30    }      //把字符分为终结符号合非终结符号       
 31    l=0;      
 32    m=0;       
 33    for(i = 0;i < line; i++)                                           
 34    {             
 35        for(j = 0;j < 30&&(p[i][j] != '\0');j++)           
 36        {                   // 非终结符号放入数组VN中                      
 37            if((p[i][j]<='z' && p[i][j]>='a')||(p[i][j]<='9' && p[i][j]>='0'))                    
 38            {                             
 39                flag = 0;                             
 40                for(t=0; VN[t] != '\0';t++)                            
 41                {                                   
 42                    if(VN[t] == p[i][j])                                  
 43                    {  
 44                        flag = 1;                                     
 45                        break;                                  
 46                    }                            
 47                }                             
 48                if(flag == 0)                            
 49                {                                   
 50                    VN[l] = p[i][j];                                 
 51                    l++;                            
 52                }                     
 53            }                      
 54            // 终结符号放入数组VT中                      
 55            if(p[i][j]<='Z' && p[i][j]>='A')                     
 56            {                             
 57                flag = 0;                             
 58                for(t = 0; t<30&&(VT[t] != '\0'); t++)                           
 59                {                                  
 60                    if(VT[t] == p[i][j])
 61                    {
 62                        flag = 1; 
 63                        break; 
 64                    }  
 65                } 
 66                if(flag==0) 
 67                { 
 68                    VT[m] = p[i][j]; 
 69                    m++; 
 70                }  
 71            } 
 72        } 
 73    } //把规则右部分分离放入数组q中
 74    count = 0; 
 75    k =0; 
 76    for(i = 0;i < line;i++) 
 77    { 
 78        for(j = 4;j < 30 && (p[i][j] != '\0');j++) 
 79        { 
 80            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'))
 81            {  
 82                q[count][k] = p[i][j]; 
 83                k++; 
 84            } 
 85            else 
 86            { 
 87                count++; 
 88                k =0; 
 89            } 
 90        }  
 91        count++; 
 92        k=0; 
 93    } //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
 94    //判断依据：q数组中每一行字符串是否相同
 95    flag = 0; 
 96    for(i = 0;i < count  i++) 
 97    { 
 98        for(j = i+1; j < count  j++) 
 99        { 
100            if(strcmp(q[i],q[j]) == 0) 
101            { 
102                flag = 1; 
103                break; 
104            } 
105        } 
106    } 
107    if(flag == 1) 
108    { 
109        printf("是非确定的有穷状态自动机，即NFA\n\n"); 
110        printf("构造的有穷状态自动机为：\n"); 
111        printf("NFA N =（K，∑，M，{S}，{Z}）\n"); 
112    } 
113    else 
114    { 
115        printf("是确定的有穷状态自动机，即DFA\n\n\n"); 
116        printf("构造的有穷状态自动机为：\n"); 
117        printf("DFA D=（K，∑，M，{S}，{Z}）\n"); 
118    } 
119    printf("其中，\nK = {S"); 
120    for(i = 0;i < 30 && (VT[i]!='\0');i++) 
121    { 
122        printf(",%c",VT[i]); 
123    } 
124    printf("}\n"); 
125    printf("∑={"); 
126    for(i = 0;i < 30 && (VN[i]!='\0');i++) 
127    { 
128        printf("%c",VN[i]); 
129    } 
130    printf("}\n"); //分离文法；
131    k = 0; 
132    count = 0; 
133    for(i = 0; i < line; i++) 
134    { 
135        j = 4; 
136        while(p[i][j] != '\0') 
137        { 
138            if(k < 4) 
139            { 
140                q[count][k] = p[i][k]; 
141                k++; 
142            } 
143            else 
144            { 
145                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')) 
146                { 
147                    q[count][k] = p[i][j]; 
148                    k++; 
149                    j++; 
150                } 
151                if(p[i][j] == '|') 
152                { 
153                    count++; 
154                    k = 0; 
155                    j++; 
156                } 
157            } 
158        } 
159        count++; 
160        k=0; 
161    } 
162    printf("\n"); 
163    //打印M后半部分
164    printf("M:\n"); 
165    l =0; 
166    while(VN[l] != '\0') 
167    { 
168        printf("M(S,%c)={",VN[l]); 
169        for(i = 0; i < 30  i++) 
170        { 
171            for(j = 4; j<30 && (q[i][j] != '\0');j++) 
172            { 
173                if(VN[l]==q[i][j]&&(q[i][j+1] == '\0')&&(q[i][j-1] == '=')) 
174                    printf("%c ",q[i][0]); 
175            } 
176        } 
177        printf("}\t"); 
178        l++; 
179    } 
180    printf("\n"); 
181    l = 0; 
182    k = 0; 
183    while(VT[k] != '\0') 
184    { 
185        l =0; 
186        while(VN[l] != '\0') 
187        { 
188            printf("M(%c,%c)={",VT[k],VN[l]); 
189            for(i = 0; i < 30  i++) 
190            { 
191                for(j = 4; j<30 && (q[i][j] != '\0');j++) 
192                { 
193                    if(VT[k] == q[i][j] && VN[l] == q[i][j+1]) 
194                        printf("%c ",q[i][0]); 
195                } 
196            }
197            printf("}\t"); 
198            l++; 
199        } 
200        k++; 
201        printf("\n"); 
202    } 
203    system("pause"); 
204 }

总结：

第一个是自己思考的，接下来不知道怎么弄，虽然做得不好，可是也是经过思考的。第二个是借鉴同学的，用来学习 的。知道利用递归来完成自动机，可是不会算法，只能依稀地写下大概的结构框架。