C++ InitStack函数代码示例

Status TopologicalOrder(ALGraph G, Stack &T) {  // 算法7.13
  // 有向网G采用邻接表存储结构，求各顶点事件的最早发生时间ve(全局变量)。
  // T为拓扑序列定点栈，S为零入度顶点栈。
  // 若G无回路，则用栈T返回G的一个拓扑序列，且函数值为OK，否则为ERROR。
  Stack S;int count=0,k;
  char indegree[40];
  ArcNode *p;
  InitStack(S);
  FindInDegree(G, indegree);  // 对各顶点求入度indegree[0..vernum-1]
  for (int j=0; j<G.vexnum; ++j)     // 建零入度顶点栈S
    if (indegree[j]==0) Push(S, j);  // 入度为0者进栈
  InitStack(T);//建拓扑序列顶点栈T
  count = 0;  
  for(int i=0; i<G.vexnum; i++) ve[i] = 0;  // 初始化
  while (!StackEmpty(S)) {
    Pop(S, j);  Push(T, j);  ++count;       // j号顶点入T栈并计数
    for (p=G.vertices[j].firstarc;  p;  p=p->nextarc) {
      k = p->adjvex;            // 对j号顶点的每个邻接点的入度减1
      if (--indegree[k] == 0) Push(S, k);   // 若入度减为0，则入栈
      if (ve[j]+p->info > ve[k])  ve[k] = ve[j]+p->info;
    }//for  *(p->info)=dut(<j,k>)
  }//while
  if (count<G.vexnum) return ERROR;  // 该有向网有回路
  else return OK;
} // TopologicalOrder

ElemType OperaType(void){
	//全数据为整型操作
	SqStack OPTR,OPND;//OPTR运算符栈，OPND运算栈
	InitStack(OPTR);
	Push(OPTR,'#');
	InitStack(OPND);
	int ch,x,a,b,theta;
	ch=getche();
	while(ch!='#' || GetTop(OPTR)!='#'){
		if(!In(ch)){
			Push(OPND,ch-48);
			ch=getche();
		}else{
			switch(Precede(GetTop(OPTR),ch)){
			case '<':
				Push(OPTR,ch);
				ch=getche();
				break;
			case '=':
				Pop(OPTR,x);
				ch=getche();
				break;
			case '>':
				Pop(OPTR,theta);
				Pop(OPND,b);
				Pop(OPND,a);
				Push(OPND,Operate(a,theta,b));
				break;
			}//switch
		}//if else
	}//while
	return GetTop(OPND);
}// OperaTyp

//编写main函数
//维护两个栈，一个为op，用来保存操作符号，一个为num，用来保存操作数
int main() {
	SqStack opStack, numStack;
	InitStack(opStack);
	InitStack(numStack);//初始化两个栈
	Push(opStack, '#');
	SElemType op, nextop;//获取弹出的操作符
	SElemType first = -1, second = -1, result = -1;//获取弹出的操作数
	SElemType ch;
	
	ch = getchar();
	GetTop(opStack, op);
	while (ch != '=' || op != '#')//如果不结束，一直获取字符
	{
		/*
			这里很关键，对数字进行移动
		*/
		if (ch >= '0'&&ch <= '9')
		{
			SElemType temp = 0;//清0
			while (ch >= '0'&&ch <= '9')
			{
				temp = temp * 10 + (ch-'0');
				ch = getchar();
			}
			Push(numStack, temp);
		}
		else
		{
			switch (compare(op, ch))
			{
				//当下一个操作字符特权高于栈顶时,直接压栈
			case -1:
				Push(opStack, ch);
				ch = getchar();//获取下一个字符
				break;
				//当特权相等时，弹栈
			case 0:
				Pop(opStack, op);
				ch = getchar();//获取下一个字符
				break;
				//当下一个操作符特权小于栈顶时，计算并压栈
			case 1:
				Pop(numStack, second);//首先获取的是第二个操作数
				Pop(numStack, first);//获取第一个操作数
				Pop(opStack, op);//获取操作符
				result = canculate(first, second, op);
				Push(numStack, result);
				//不能将操作数压栈，要进行新一轮比较
				//Push(opStack, ch);
				break;
			}
		}

		GetTop(opStack, op);
	}
	GetTop(numStack, result);//获取栈中的第一个元素
	printf("%d\n", result);
	return 0;
}

/**
 * 习题3.21，将中序表达式写成逆波兰式
 */
void RPN(SqStack &S1, SqStack &S2)
{
	char c;
	int temp;
	InitStack(S1);		//存储临时运算符
	InitStack(S2);		//存储逆波兰式
	Push(S1, '#');
	c = getchar();
	while (c != '#' || !StackEmpty(S1)) {
		if (!In(c)) {		//不是运算符，是操作数
			Push(S2, c);
			c = getchar();	//读入下一个字符
		} else {		//是运算符
			if ('(' == c) {
				Push(S1, c);
				c = getchar();	//读入下一个字符
			} else if (')' == c) {
				while (GetTop(S1) != '(') {
					Pop(S1, temp);
					Push(S2, temp);
				}
				if (GetTop(S1) == '(')
					Pop(S1, temp);
				c = getchar();
			} else {
				switch (Precede(GetTop(S1), c)) {
				case '<':
					Push(S1, c);
					c = getchar();
					break;
				case '>':
					while ('>' == Precede(GetTop(S1), c)) {
						Pop(S1, temp);
						Push(S2, temp);
					}
					Push(S1, c);
					c = getchar();
					break;
				}
			}
		}
		if ('#' == c) {
			while ('#' != GetTop(S1)) {
				Pop(S1, temp);
				Push(S2, temp);
			}
			Pop(S1, temp);
		}
	
	}
}

int Start(SqStack &s,int maze[][XSIZE])
{
	SqStack a;
	SElemType e;
	InitStack(a);
	InitStack(s);
	Step(a,maze);
	while (a.top>a.base)
	{
		Pop(a,e);
		Push(s,e);
	}
	return OK;
}

Status TopologicalOrder(ALGraph &G,SqStack &T)
{
	int i,k,count = 0;
	int indegree[MAX_VERTEX_NUM];//入度数组,存放各顶当前入度数
	SqStack S;//零入度栈 也可用队列代替
	ArcNode *p;
	FindIndegree(G,indegree);//对G的各顶点求入度indegree[]
	InitStack(S);//初始化零入度栈
	for(i = 0; i < G.vexnum; ++i)//将零入度的顶点序号入栈
	{
		if (!indegree[i])//入度为零
		{
			Push(S,i);
		}
	}
	InitStack(T);
	for (i = 0; i < G.vexnum; ++i)//初始化ve为0
	{
		G.vertices[i].data.ve = 0;
	}
	while (!StackEmpty(S))//当栈不空时
	{
		Pop(S,i);//出栈一个零入度顶点的序号,并将其赋值给i
		Visit(G.vertices[i].data);
		Push(T,i);//i号顶点入逆拓扑排序栈T(栈底元素为拓扑排序的第1个元素)
		++count;//已输出顶点数+1
		for (p = G.vertices[i].firstarc; p; p = p->nextarc)//对i号的顶点的每个邻接顶点
		{
			k = p->data.adjvex;//其序号为K
			if (!(--indegree[k]))//k的入度减1,若减为0,则将k入栈S
			{
				Push(S,k);
			}
			if ((G.vertices[i].data.ve + p->data.info->weight) > G.vertices[k].data.ve)
			{//顶点i事件的最早发生时间 + <i,k>的权值  > 顶点j事件的最早发生时间
				G.vertices[k].data.ve = G.vertices[i].data.ve + p->data.info->weight;
				//顶点j事件的最早发生时间 = 顶点i事件的最早发生时间 + <i,k>的权值
			}
		}
	}
	if (count < G.vexnum)//零入度顶点栈S已空,图还有顶点未输出
	{
		printf("此有向图有回路！\n");
		return ERROR;
	}
	else
	{
		return OK;
	}
}

int main()
{
    ElemType c, temp;
    sqStack s, r;
    int len, i, t, sum=0;

    InitStack(&s);
    InitStack(&r);

    printf("Please input a bin number, # to stop!\n");
    scanf("%c", &c);

    while(c != '#')
    {
        Push(&s, c);
        scanf("%c", &c);
    }

    getchar();  // 把'\n'从缓冲区去掉  也可用fflush(stdin)

    len = StackLen(s);
    printf("The current length of the stack is: %d\n", len);

    for(i=0; i<len; i++)
    {
        Pop(&s, &c);
        sum += (c-48) * pow(2, i%3);


        if((i+1)%3 == 0 || i+1 == len)
        {

            Push(&r, sum);
            sum = 0;
        }
    }

    len = StackLen(r);
    printf("The current length of the new stack is: %d\n", len);

    printf("Convert to Oct is: ");
    for(i=0; i<len; i++)
    {
        Pop(&r, &c);
        printf("%o", c);
    }

    printf("\n");
    return 0;
}

int EvalExpres(void)    /* 表达式求解函数*/
{
	int a,b,i=0,s=0;
	char c[80],r;
	InitStack(&StackR);
	Push(&StackR,'#');
	InitStack(&StackD);
	printf(" 请输入表达式并以‘#’结束：");
	gets(c);
	while(c[i]!='#' || GetTop(&StackR)!='#')
	{      
		if(!In(c[i]))   /* 判断读入的字符不是运算符 是则进栈*/
		{ if(c[i] >= '0' && c[i] <= '9')
		{
			s += c[i]-'0';      
			while(!In(c[++i]))    /*此处实现的功能为当输入的数字为多位数时*/
			{ s*=10;
			s += c[i]-'0';
			}
			Push(&StackD,s+'0');
			s = 0;
		}
		else
		{
			printf("你输入的表达式有误!\n");
			return 0;
		}            
		}
		else        
			switch(Proceed(GetTop(&StackR),c[i])) /* 此函数用来比较读取的运算符和栈顶运算符的优先级*/
		{
			case '<':   /* 栈顶的元素优先级高*/
				Push(&StackR,c[i]);
				i++;
				break;
			case '=':   /* 遇到匹配的小括号时则去掉他*/
				Pop(&StackR);
				i++;
				break;
			case '>': /* 栈顶的优先级低则出栈并将结果写入栈内*/
				r = Pop(&StackR);
				a = Pop(&StackD)-'0';
				b = Pop(&StackD)-'0';
				Push(&StackD,Operate(a,r,b)) ;
				break;                
		}
	}
	return (GetTop(&StackD)-'0');   /* 返回运算结果*/
}

void parenthesisMatch() {
	char s[100];
	printf("please input the string which consists of paretheses:\n");
	scanf("%s", s);
	SqStack S;
	InitStack(S);
	for (int i = 0; s[i] != '\0'; i++) {
		if (s[i] == '[' || s[i] == '{' || s[i] == '(') {
			ElemType e;
			e.key = s[i];
			Push(S, e);
		}
		else if (s[i] == ']' || s[i] == '}' || s[i] == ')') {
			ElemType e;
			Pop(S, e);
			if (e.key != s[i] - 1 && e.key != s[i] - 2) {
				printf("the parentheses are not matching.\n");
				return;
			}
		}
	}
	if (StackEmpty(S)) {
		printf("the parentheses are matching.\n");
	}
	DestroyStack(S);
}

void conversion() {
	SqStack S;
	InitStack(S);
	int N;
	printf("input the decimal number:\n");
	scanf_s("%d", &N);
	printf("input the the number system after conversion:\n");
	int d;
	scanf_s("%d", &d);
	while (N) {
		ElemType e;
		e.key = N % d;
		Push(S, e);
		N = N / d;
	}
	PrintStack(S);
	printf("the conversion result is:\n");
	while (!StackEmpty(S)) {
		ElemType e;
		Pop(S, e);
		printf("%d", e.key);
	}
	printf("\n");
	PrintStack(S);
	DestroyStack(S);
}

void lineEdit() {
	SqStack S;
	InitStack(S);
	char ch;
	ch = getchar();
	ElemType e;
	while (ch != EOF) {
		while (ch != EOF && ch != '\n') {
			switch (ch) {
			case '#': 
				Pop(S, e);
				break;
			case '@':
				ClearStack(S);
				break;
			default:
				e.key = ch;
				Push(S, e);
				break;
			}
			ch = getchar();
		}

		PrintStack(S);
		ClearStack(S);
		if (ch != EOF) ch = getchar();
	}
	DestroyStack(S);
}

int main()
{
    ElemType c;
    sqstack s;
    int len, i, sum=0;

    InitStack(&s);

    printf("Please input binary number, use symbole '#' to mark the end: ");
    scanf("%c", &c);
    while(c != '#'){
        Push(&s,c);
        scanf("%c",&c);
    }

    getchar(); // remove '\n' from cache

    len = StackLen(s);
    printf("Stack length is: %d\n",len);

    for(i=0;i<=len;i++){
        Pop(&s,&c);
        sum = sum+(c-48)*pow(2,i);
    }

    printf("The decimal number is: %d", sum);

    return 0;
}

void LineEdit(){
	// P50 算法3.2 行编辑程序
	// #:退格 @:退行
	// 利用字符栈S，从终端接收一行并传送至调用过程的数据区
	SqStack S;
	InitStack(S);
	SElemType e;
	char ch = getchar();
	while (ch != EOF){
		while (ch != EOF && ch != '\n'){
			switch (ch){
				case'#': 
					Pop(S, e);
					break;
				case'@':
					ClearStack(S);
					break;
				default:
					Push(S, ch);
					break;
			}
			ch = getchar();	//从终端接收下一个字符
		}
		StackTraverse(S, Visit_Display_Char);
		ClearStack(S);
		if (ch != EOF) ch = getchar();
	}
	DestroyStack(S);
}// LineEdit

bool kuohao_pipei(char *str) {
	/*
	×¼±¸Ò»¸öÕ»£¬ÓÃÓÚ´æ·ÅɨÃèÓöµ½µÄ×óÀ¨ºÅ
	´Ó×óÏòºóɨÃèÿһ¸ö×Ö·û{
	  Èç¹ûÓöµ½µÄÊÇ×óÀ¨ºÅ£¬ÔòÈëÕ»
	  Èç¹ûÓöµ½µÄÊÇÓÒÀ¨ºÅ£¬Ôò
	     °ÑÕ»¶¥×Ö·ûºÍµ±Ç°×Ö·û±È½Ï
	     ÈôÆ¥Å䣬Ôòµ¯³öÕ»¶¥×Ö·û£¬¼ÌÐøÏòǰɨÃè
	     Èô²»Æ¥Å䣬³ÌÐò·µ»Ø²»Æ¥Åä±êÖ¾
	}
	µ±ËùÓÐ×Ö·û¶¼É¨ÃèÍê±Ï£¬Õ»Ó¦µ±Îª¿Õ
	*/
	SqStack S; InitStack(S);

	for (int i = 0; str[i] != '\0'; i++) {
		if (isLeftKuohao(str[i])) {
			Push(S, str[i]);
		}
		else if (isRightKuohao(str[i])) {
			char ch;
			Top(S, ch);
			if (isMatch(ch, str[i]) == false)
				return false;
			else {
				Pop(S, ch);
			}
		}
	}

	if (isEmpty(S)) return true;
	return false;	
}

int main()
{
	// Let's make a stack -- Always a good idea initialize variables to some known value
	SStack stack = {0};
	int counter = 0; // A counter for our "for loops"
	
	InitStack(&stack,10);


	// Let's fill the stack with numbers 0 - 9.
	// When this for loop is done the stack will be FULL.
	for(counter = 0; counter < 10; counter++)
		Push(&stack,counter);
	
	if(Push(&stack,22))
		printf("This is we bad! We just pushed an element onto a full stack.\n\n");

	else
		printf("This is good!  We were denied while trying to \"push\" onto a full stack.\n\n");


	// We'll print out the numbers.  You'll see there in LIFO order (The last one we
	// pushed on, is the first one we'll pop off, etc).
	for(counter = 0; counter < 10; counter++)
		printf("We just popped off #%d\n",Pop(&stack));
		
	
	if(FreeStack(&stack))
		printf("\nYeah the memory was freed successfully!\n\n");
	else
		printf("\nOh NO!  An error occurred while trying to free the memory");

	return 0;
}