#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]);
}
//把字符分为终结符和非终结符
l=0;
m=0;
for(i=0;i<line;i++)
{
for(j=0;j<30&&(p[i][j]!=‘\0‘);j++) { //非终结符放入数组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++;
}
}
//终结符放在数组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,E(总和的意思),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 N=(K,E(总和的意思),M,{S},{Z})\n");
}
printf("其中,\nK={S");
for(i=0;i<30&&(VT!=‘\0‘);i++)
{
printf(",%c",VT[i]);
}
printf("}\n");
printf("E={");
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]==‘l‘)
{
count++;
k=0;
j++;
}
}
}
count++;
k=0;
}
printf("\n");
//打印M后部分
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");
}
1203有穷自动机
时间: 2024-12-14 12:27:01
1203有穷自动机的相关文章
1203 有穷自动机的构造
这是一个未完成的程序 #include<stdio.h> #include<string.h> char a[20][10]; char Vn[26]; char Vt[64]; int M=0; int N=0; int m=0; int flag=-1; void initial() //³õʼ»¯ { for(int b=0;b<26;b++) Vn[b]='#'; for(int c=0;c<64;c++) Vt[c]='#'; for(int d=0;d&l
1203 有穷自动机
#include<stdio.h> #define MAX 100 typedef struct //构造一个邻接表 用于存储NFA { char name; char line[MAX]; }node; void tran(){ //专门做语句的转换操作 } void automata(char R[],int i){ int j = 0; int n = i; while(R[j] != '#'){ if(R[i] == '(') { printf("//在这里是做把
1203正规式转换为有穷自动机
1 #include<stdio.h> 2 #include <ctype.h> 3 #define ok 1 4 #define error 0 5 #define MAXREGLUARLONG 40 6 #define MAXSTATELONG 40 7 #define MAXCAHRSLONG 40 8 typedef int state; 9 int iCurrentState=0; //初态以1开始 10 int iPreState=0; 11 int iLastFork
TLS 协议所定义的严重错误代码是 10。Windows SChannel 错误状态是 1203
windows 2012 操作系统下面报36888/36887. 生成了一个严重警告并将其发送到远程终结点.这会导致连接终止.TLS 协议所定义的严重错误代码是 10.Windows SChannel 错误状态是 1203. 从远程终点接收到一个严重警告.TLS 协议所定义的严重警告代码为 48. 其实schannel的事件录入是分成4个等级的: Logging options The default value for Schannel event logging is 0×0000 in W
TOJ 1203: Number Sequence
1203: Number Sequence Time Limit(Common/Java):1000MS/10000MS Memory Limit:65536KByte Total Submit: 957 Accepted:208 Description A number sequence is defined as follows: f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7. Gi
hzau 1203 One Stroke
1203: One Stroke Time Limit: 2 Sec Memory Limit: 1280 MBSubmit: 264 Solved: 56[Submit][Status][Web Board] Description There is a complete binary tree which includes n nodes. Each node on the tree has a weight w, each edge on the tree is directed fr
ACdream 1203 - KIDx&#39;s Triangle(解题报告)
KIDx's Triangle Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others) Submit Statistic Next Problem Problem Description One day, KIDx solved a math problem for middle students in seconds! And than he created this problem. N
hdoj 1203 I NEED A OFFER! 【另类01背包】【概率背包】
题意:... 策略:动态规划. 因为是求至少能得到一个offer的概率,那我们可以反着求,求得不到一个offer的概率,最后用1减去就好了. 代码: #include<string.h> #include<stdio.h> double dp[10010]; struct node{ int a; double b; }s[10010]; int main() { int n, m, i, j; while(scanf("%d%d", &n, &
杭电 1203 I NEED A OFFER!(01背包)
http://acm.hdu.edu.cn/showproblem.php?pid=1203 I NEED A OFFER! Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 15759 Accepted Submission(s): 6259 Problem Description Speakless很早就想出国,现在他已经考完了