PTA数据结构与算法题目集（中文）4-7 在一个数组中实现两个堆栈 (20分)

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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 1e8typedef 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：

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

输出样例：

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

程序代码：

Stack CreateStack( int MaxSize ){Stack sq=(struct SNode*)malloc(sizeof(struct SNode));sq->Data=(ElementType *)malloc(sizeof(ElementType)*MaxSize);sq->MaxSize=MaxSize;sq->Top1=-1;sq->Top2=MaxSize;return sq;}bool Push( Stack S, ElementType X, int Tag ){if(S->Top1+1==S->Top2){printf("Stack Full\n");return false;}if(Tag==1){S->Top1++;S->Data[S->Top1]=X;}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 1 Empty\n");return ERROR;}else{return S->Data[S->Top1--];//先返回S->Data[S->Top1] 然后 S->Top1自减 }}else{if(S->Top2==S->MaxSize){printf("Stack 2 Empty\n");return ERROR;}else{return S->Data[S->Top2++];//先返回S->Data[S->Top2] 然后 S->Top2自加 }}}

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