1.问题的描述：有一个魔王总是用自己一种非常精练而抽象的语言讲话，没有人能听懂，但他的语言可以逐步的转换成人的语言，他的语言由以下两种转换规则由人的语言逐步抽象上去的：
2.基本要求：实现用下述两条具体规则和上述规则（2）编写魔语解释系统。设大写字母表示魔王语言的词汇；小写字母表示人的词汇；希腊字母表示可以用大写字母或小写字母代换的变量。魔王语言可以含人的词汇。（1）B tAdA；（2）A sae.
#include
#include
#include
#include
#define LEN 20
#define L_size 100
struct
{
char rA[20];
char ra[20];
}rule[LEN];
struct
{
char mA[20];
char ma[20];
}mean[LEN];
//栈的操作
typedef struct node
{
char data;
struct node * next;
}LinkStackNode, * LinkStack;
//初始化栈
void InitStack(LinkStack * top)
{
(* top) = (LinkStack)malloc(sizeof(LinkStackNode));
(* top) -> next = NULL;
}
//入栈
void Push(LinkStack top, char a)
{
LinkStackNode * temp;
temp = (LinkStackNode *)malloc(sizeof(LinkStackNode));
temp -> data = a;
temp -> next = top -> next;
top -> next = temp;
}
//出栈
void Pop(LinkStack top, char *x)
{
LinkStackNode * temp;
temp = top -> next;
top -> next = temp -> next;
*x = temp -> data;
free(temp);
}
//队列的操作
typedef struct Node
{
char data;
struct Node * next;
}LinkQueueNode;
typedef struct
{
LinkQueueNode * front;
LinkQueueNode * rear;
}LinkQueue;
//队列初始化
void InitQueue(LinkQueue *Q)
{
Q -> front = (LinkQueueNode *)malloc(sizeof(LinkQueueNode));
if(Q->front != NULL)
{
Q -> rear = Q -> front;
Q -> front -> next = NULL;
}
}
//入队
void EnterQueue(LinkQueue *Q, char x)
{
LinkQueueNode * NewNode;
NewNode=(LinkQueueNode *)malloc(sizeof(LinkQueueNode));
if(NewNode != NULL)
{
NewNode -> data = x;
NewNode -> next = NULL;
Q -> rear -> next = NewNode;
Q -> rear = NewNode;
}
}
//出队
void DeleteQueue(LinkQueue *Q, char *x)
{
LinkQueueNode * p;
p = Q -> front -> next;
Q -> front -> next = p -> next;
if(Q -> rear == p)
Q -> rear = Q -> front;
*x = p -> data;
free(p);
}
//读规则文件
void read_rulefile() //读取文件
{
FILE * fp;
char filename[LEN];
int i = 0;
printf("Please enter the name of the file to open rules:");//读取文本文件(rule.txt), 将具体规则的对应关系读入数组
fgets(filename, LEN, stdin);
fp = fopen("rule.txt", "rt");
if(NULL == fp)
{
printf("\nFailed to open file，%sFile may not exist\n",filename);
printf("EXIT\n");
exit(1);
}
while(fscanf(fp, "%s %s", rule[i].rA, rule[i].ra) != EOF)
{
printf("%s %s\n", rule[i].rA, rule[i].ra);
i++;
}
fclose(fp);
return;
}
void read_meanfile()
{
FILE * fp;
char filename[LEN];
int i = 0;
printf("Please enter the name of the file to open the meaning of:");//读取文本文件(mean.txt), 将小写字母及其对应的含义读入数组
fgets(filename, LEN, stdin);
fp = fopen("mean.txt", "rt");
if(NULL == fp)
{
printf("\nFailed to open file，%sFile may not exist\n", filename);
printf("EXITI\n");
exit(1);
}
while(fscanf(fp, "%s %s", mean[i].mA, mean[i].ma) != EOF)
{
printf("%s %s\n", mean[i].mA, mean[i].ma);
i++;
}
fclose(fp) ;
return;
}
//写文件操作
void save_result(char L[]) //把结果写入result.txt文档中
{
FILE * fp;
char filename[LEN];
int i = 0, j;
printf("\nPlease enter the file to be saved file name:");
scanf("%s", filename);
if((fp = fopen(filename, "wt")) == NULL)
{
printf("Write file error, press any key to exit!");
exit(1);
}
while(L[i] != '\0')
{
for(j = 0; j < 20; j++)
{
if(L[i] == mean[j].mA[0])
fprintf(fp, "%s", mean[j].ma);
}
i++;
}
printf("File is saved successfully, press any key to return!\n");
fclose(fp);
}
//去括号
/*思想：当没遇到闭括号时，一直压栈（top栈）。一旦遇到闭括号，首先找到与这个闭括号最近的匹配的开括号找到这两个括号“（” “）”之间的第一个字符利用队列（Q 入队）以及（top栈出栈）完成转换，再把转换后的队列，添加到（top栈中）把剩余的字符原样添加到top栈中在把top栈逆置到top1栈中再利用top1栈出栈，给L重新赋值。递归，看L中是否还有括号*/
void tackle_2(LinkStackNode * top, LinkStackNode * top1, LinkQueue * Q, char L[])
{
int i = 0, j;
char * a;
char first;
a = (char *)malloc(sizeof(char));
InitStack(&top);
InitQueue(Q); //辅助用来输出括号内的内容
InitStack(&top1);
while(L[i] != ')')
{
Push(top, L[i]);//当不等于‘）’时，往top栈中压。
i++;
}
if(L[i] == ')') //当等于'）'时
{
j = i;
while(L[j] != '(')
{
j--;
if(L[j] == '(')
{
j++;
break;
}
}
first = L[j];//找到当前‘（ ’内的第一个字符
for(;j < i - 1; j++)
{
EnterQueue(Q, first);
Pop(top, a);
EnterQueue(Q, *a);
}
Pop(top, a); //这个是括号内的第一个字符
EnterQueue(Q, *a);
Pop(top, a); //把‘（’删掉
while(Q -> front -> next != NULL)
{
DeleteQueue(Q,a);
Push(top,*a);
}
} //if
i++; //跳过‘）’
while(L[i] != '\0')
{
Push(top, L[i]);
i++;
}
while(top -> next != NULL)
{
Pop(top, a);
Push(top1, *a);
}
i = 0;
while(top1 -> next != NULL)
{
Pop(top1,a);
L[i] = *a;
i++;
}
L[i] = '\0';
i = 0;
j = 0;
while(L[i] != '\0')
{
i++;
if(L[i] == '(' || L[i] == ')')
j++;
}
if(j == 0)
return;
else
tackle_2(top, top1, Q, L);
}
//实现提示:处理规则形式(1)问题并没有规定具体规则的数量.比如增加一条具体规则C→eB那么对魔王语言就要多处理一次,因此处理规则形式(1)时需要递归处理,当语言中没有大写字母后递归结束。
void tackle_1(char L[], LinkQueue * Q)
{
int i = 0, j, z;
int k = 0;
char * a;
a = (char *)malloc(sizeof(char));
while(L[i] != '\0')
{
z = 0;
for(j = 0; j < 20; j++)
{
if(L[i] == rule[j].rA[0])
{
k = 0;
z = 1;
while(rule[j].ra[k] != '\0')
{
EnterQueue(Q, rule[j].ra[k]);
k++;
}
}
}
if(z == 1)
{
i++;
}
else
{
EnterQueue(Q, L[i]);
i++;
}
}
i = 0;
while(Q -> front -> next != NULL)
{
DeleteQueue(Q, a);
L[i] = (*a);
i++;
}
L[i] = '\0';
i = 0;
j = 0;
while(L[i] != '\0')
{
if(L[i] >= 'A' && L[i] <= 'Z')
j++;
i++;
}
if(j == 0)
return;
else
tackle_1(L, Q);////递归调用
}
void tackle_12(char L[])
{
int i = 0;
int j;
while(L[i] != '\0')
{
for(j = 0; j < 20; j++)
{
if(L[i] == mean[j].mA[0])
printf("%s", mean[j].ma);
}
i++;
}
printf("\n");
}
int main(int argc, char *argv[])
{
int j = 0;
read_rulefile();
read_meanfile();
char L[L_size]; //魔王说的话
printf("Please enter the devil king language:");
scanf("%s", L);
LinkStackNode * top, * top1;
top = (LinkStackNode * )malloc(sizeof(LinkStackNode));
top1 = (LinkStackNode * )malloc(sizeof(LinkStackNode));
LinkQueue *Q;
Q = (LinkQueue *)malloc(sizeof(LinkQueue));
tackle_2(top, top1, Q, L);
printf("Devil king to say what is：%s", L);
tackle_1(L, Q);
printf("\nDevil to say what is（Convert after lowercase）：%s\n", L);
tackle_12(L);
printf("Devil to say mean：%s\n", mean[j].ma);
tackle_12(L);
save_result(L);
return EXIT_SUCCESS;
}
