在一个数组中实现两个堆栈

6-7 在一个数组中实现两个堆栈(20 分)

本题要求在一个数组中实现两个堆栈。

函数接口定义:

Stack CreateStack( int MaxSize );
bool Push( Stack S, ElementType X, int Tag );
ElementType Pop( Stack S, int Tag );

其中Tag是堆栈编号,取1或2;MaxSize堆栈数组的规模;Stack结构定义如下:

typedef int Position;
struct SNode {
    ElementType *Data;
    Position Top1, Top2;
    int MaxSize;
};
typedef struct SNode *Stack;

注意:如果堆栈已满,Push函数必须输出“Stack Full”并且返回false;如果某堆栈是空的,则Pop函数必须输出“Stack Tag Empty”(其中Tag是该堆栈的编号),并且返回ERROR。

裁判测试程序样例:

#include <stdio.h>
#include <stdlib.h>

#define ERROR 1e8
typedef int ElementType;
typedef enum { push, pop, end } Operation;
typedef enum { false, true } bool;
typedef int Position;
struct SNode {
    ElementType *Data;
    Position Top1, Top2;
    int MaxSize;
};
typedef struct SNode *Stack;

Stack CreateStack( int MaxSize );
bool Push( Stack S, ElementType X, int Tag );
ElementType Pop( Stack S, int Tag );

Operation GetOp();  /* details omitted */
void PrintStack( Stack S, int Tag ); /* details omitted */

int main()
{
    int N, Tag, X;
    Stack S;
    int done = 0;

    scanf("%d", &N);
    S = CreateStack(N);
    while ( !done ) {
        switch( GetOp() ) {
        case push: 
            scanf("%d %d", &Tag, &X);
            if (!Push(S, X, Tag)) printf("Stack %d is Full!\n", Tag);
            break;
        case pop:
            scanf("%d", &Tag);
            X = Pop(S, Tag);
            if ( X==ERROR ) printf("Stack %d is Empty!\n", Tag);
            break;
        case end:
            PrintStack(S, 1);
            PrintStack(S, 2);
            done = 1;
            break;
        }
    }
    return 0;
}

/* 你的代码将被嵌在这里 */

输入样例:

5
Push 1 1
Pop 2
Push 2 11
Push 1 2
Push 2 12
Pop 1
Push 2 13
Push 2 14
Push 1 3
Pop 2
End

输出样例:

Stack 2 Empty
Stack 2 is Empty!
Stack Full
Stack 1 is Full!
Pop from Stack 1: 1
Pop from Stack 2: 13 12 11

ANS:

Stack CreateStack( int MaxSize )
{
  Stack s = (Stack )malloc(sizeof(struct SNode));
  s->MaxSize=MaxSize;
  s->Data=(ElementType *)malloc(MaxSize*sizeof(ElementType));
  s->Top1=-1;
  s->Top2=MaxSize;
  return s;
}
bool Push( Stack S, ElementType X, int Tag )
{
  if(S->Top2-S->Top1==1)
  {
    printf("Stack Full\n");
    return 0;
  }
  if(Tag==1)
  {
    S->Data[++(S->Top1)]=X;
  }
  else 
  {
    S->Data[--(S->Top2)]=X;
  }
  return 1;
}
ElementType Pop( Stack S, int Tag )
{
  if(S->Top1==-1&&Tag==1)
  {
    printf("Stack %d Empty\n",Tag);
    return ERROR;
  }
  if(S->Top2==S->MaxSize&&Tag==2)
  {
    printf("Stack %d Empty\n",Tag);
    return ERROR;
  }
  return Tag==1?S->Data[(S->Top1)--]:S->Data[(S->Top2)++];
}

猜你喜欢

转载自blog.csdn.net/dilly__dally/article/details/81274644
今日推荐