PTA資料結構與演算法題目集（中文）4-7 在一個數組中實現兩個堆棧 (20分)

本題要求在一個數組中實現兩個堆棧。 函數介面定義：

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

其中Tag是堆棧編號，取1或2；MaxSize堆棧數組的規模；Stack結構定義如下：

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

注意：如果堆棧已滿，Push函數必須輸出“Stack Full”並且返回false；如果某堆棧是空的，則Pop函數必須輸出“Stack Tag Empty”（其中Tag是該堆棧的編號），並且返回ERROR。 裁判測試程式範例：

#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;}/* 你的代碼將被嵌在這裡 */

輸入範例：

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

輸出範例：

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

程式碼：

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] 然後 S->Top1自減 }}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] 然後 S->Top2自加 }}}