3-2(*) 在一个数组中实现两个堆栈 (10 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
结尾无空行
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
结尾无空行
/* 你的代码将被嵌在这里 */
输入样例:
4
Del
Add 5
Add 4
Add 3
Del
Del
Add 2
Add 1
Add 0
Add 10
End
结尾无空行
输出样例:
Queue Empty
5 is out
4 is out
Queue Full
3 2 1 0
结尾无空行
原代码(有错误):
Stack CreateStack( int MaxSize ) { Stack s=(Stack)malloc(sizeof(struct SNode)); s->Top1=-1; s->Top2=MaxSize; s->Data=(ElementType*)malloc(sizeof(ElementType)*MaxSize); s->MaxSize=MaxSize; return s; } bool Push( Stack S, ElementType X, int Tag ) { if(Tag==1){ if(S->Top2-S->Top1==1){ printf("Stack Full\n"); return false; } //*(S->Data)=X; S->Top1++; S->Data[S->Top1]=X; return true; } else{ if(S->Top2==0){ printf("Stack Full\n"); return false; } (S->Top2)--; //*((S->Data)+(S->Top2)-(S->Top1)-1)=X; S->Data[S->Top2]=X; } return true; } ElementType Pop( Stack S, int Tag ){ if(Tag==1){ if(S->Top1==-1){ printf("Stack %d Empty\n",Tag); return ERROR; } /* int x=S->Data[S->Top1]; S->Data[S->Top1]=0; S->Top1--; return x;*/ return S->Data[S->Top1++]; } else{ if((S->Top2)==(S->MaxSize)-1){ printf("Stack %d Empty\n",Tag); return ERROR; } /* int x=S->Data[S->Top2]; S->Data[S->Top2]=0; S->Top2++; return x;*/ return S->Data[S->Top2++]; //*((S->Data)+(S->Top2)-(S->Top1)-1)=0; } }
代码里面两个致命的错误:1.栈满的标志应该是Top2-Top1=1;2栈2空的标志是Top2=MaxSize;
题目正式分析:
本题属于基于动态顺序存储结构的堆栈的实现,实际上我们在解决栈的空间溢出的问题时常常使两个栈共享空间,一个栈的栈尾在存储空间开头,另一个的栈尾在存储空间末尾。
Stack CreateStack( int MaxSize ) { Stack s=(Stack)malloc(sizeof(struct SNode)); s->Top1=-1; s->Top2=MaxSize; s->Data=(ElementType*)malloc(sizeof(ElementType)*MaxSize); s->MaxSize=MaxSize; return s; } bool Push( Stack S, ElementType X, int Tag ) { if(S->Top2-S->Top1==1){ printf("Stack Full\n"); return false; } if(Tag==1){ S->Top1++; S->Data[S->Top1]=X; return true; } else{ S->Top2--; S->Data[S->Top2]=X; return true; } } ElementType Pop( Stack S, int Tag ){ if(Tag==1){ if(S->Top1==-1){ printf("Stack %d Empty\n",Tag); return ERROR; } return S->Data[S->Top1--]; } else{ if((S->Top2)==(S->MaxSize)){ printf("Stack %d Empty\n",Tag); return ERROR; } return S->Data[S->Top2++]; } }