[基本要求]

用下述两条具体规则和上述规则形式(2)实现。设大写字母表示魔王语言的词汇；小写字母表示人的语言词汇；希腊字母表示可以用大写字母或小写字母代换的变量。魔王语言可含人的词汇。

(1)B －> tAdA

(2)A －> sae

[测试数据]

B(ehnxgz)B解释成tsaedsaeezegexenehetsaedsae

【实现】将需要解释的魔王语言当成一个字符数组，入栈1。再依次出栈，入栈2，依次处理顶端字符，若是开括号逐一出栈入队列，直至闭括号出栈。再逐一出队列，按照规则2解释，再重新入栈2。出栈2，按照规则1解释。

#include

#include

#include

#include

typedef struct Node //节点

{

char data;

struct Node *next;

}Node,*NodePtr;

typedef struct Stak //栈链表

{

NodePtr top;

NodePtr bot;

int count;

}Stak;

//初始化栈

void InitStak(Stak *ts)

{

ts->top = (NodePtr)malloc(sizeof(Node));

if (NULL == ts->top)

{

printf("分配失败");

exit(-1);

}

ts->bot = ts->top;

ts->count = 0;

ts->top->next = NULL;

}

//入栈

void Push(Stak *S, char e)

{

NodePtr pnew = (NodePtr)malloc(sizeof(Node));

pnew->data = e;

pnew->next = S->top;

S->top = pnew;

S->count++;

}

//出栈

char Pop(Stak *S)

{

NodePtr p;

if (S->bot == S->top)

{//空时返回1 删除失败

exit(1);

}

p = S->top;

char e = p->data;

S->top = S->top->next;

free(p);

S->count--;

return e;

}

//打印栈

void prin(Stak *S)

{

NodePtr p;

p = S->top;

while (p != S->bot)

{

printf("%c\n",p->data);

p = p->next;

}

}

bool EmptyStack(Stak *S)

//判断是否空栈

{

if (S->count == 0)

return 1;

return 0;

}

typedef struct QNode //队列节点

{

char data;

struct QNode *next;

}QNode,*QNodePtr;

typedef struct LinkQueue //队列链表

{

QNodePtr front, rear;

}Queue;

//队列初始化

void InitQue(Queue *Q)

{

QNodePtr p = (QNodePtr)malloc(sizeof(QNode)); //p为头节点

p->data = NULL;

p->next = NULL;

Q->front = p;

Q->rear = p;

}

//入队列

void EnQue(Queue *Q, char e)

{

QNodePtr p = (QNodePtr)malloc(sizeof(QNode));

p->data = e;

p->next = NULL;

Q->rear->next = p;

Q->rear = p;

}

char DeQue(Queue *Q)

//出队列

{

QNodePtr p;

char c;

if (Q->front == Q->rear)

{//空时

exit(1);

}

p = Q->front->next;

c = Q->front->next->data;

Q->front->next = p->next;

if (p == Q->rear)

{

Q->rear = Q->front;

}

free(p);

return c;

}

void pri(Queue *Q)

//打印队列

{

QNodePtr p;

p = Q->front;

if (p != NULL)

{

while (p != Q->rear)

{

printf("%c", p->next->data);

p = p->next;

}

}

}

bool EmptyQue(Queue *Q)

{

if (Q->front == Q->rear)

return 1;

return 0;

}

void Reverse(char M[], Stak *S)

{

int i;

int len = strlen(M);

int l = 0, r = 0;

for (i =0; i

{

Push(S, M[i]);

if (M[i] == '(')

l++;

if (M[i] == ')')

r++;

}

if (l != r)

exit(1);

}

void EnA(Queue *Q)

//规则1

{

EnQue(Q, 's');

EnQue(Q, 'a');

EnQue(Q, 'e');

}

void EnB(Queue *Q)

//规则1

{

EnQue(Q, 't');

EnA(Q);

EnQue(Q, 'd');

EnA(Q);

}

void Fpri(Queue *Q)

{

char c;

while (!EmptyQue(Q))

{

c = DeQue(Q);

switch (c)

{

case 't':printf("天"); break;

case 'd':printf("地"); break;

case 's':printf("上"); break;

case 'a':printf("一只"); break;

case 'e':printf("鹅"); break;

case 'z':printf("追"); break;

case 'g':printf("赶"); break;

case 'x':printf("下"); break;

case 'n':printf("蛋"); break;

case 'h':printf("恨"); break;

default: printf("Error");

}

}

}

void Tran(Stak *Sbe,Stak *Saf,Queue *Q )

{

char c;

char d;

int i=0;

while (Sbe->count != 0)

{

c = Pop(Sbe);

if (c == ')')

i = Saf->count;

else if (c == '(')

{

int j = Saf->count;

while ( j > i )

{

d = Pop(Saf);

EnQue(Q, d);

j--;

}

char e = Q->front->next->data;

DeQue(Q);

while (!EmptyQue(Q))

//规则2

{

char dl = DeQue(Q);

Push(Saf, e);

Push(Saf, dl);

}

Push(Saf, e);

}

else

Push(Saf, c);

}

while (Saf->count>0)

{

char en;

en = Pop(Saf);

if (en == 'A')

EnA(Q);

else if (en == 'B')

EnB(Q);

else

EnQue(Q,en);

}

Fpri(Q);

}

int main()

{

Stak Sf,St;

InitStak(&Sf);

InitStak(&St);

Queue Qu1;

InitQue(&Qu1);

char M[] = "B(ehnxgz)B";

Reverse(M, &Sf);

Tran(&Sf, &St, &Qu1);

getchar();

return 0;

}