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PTA 資料結構與演算法題目集(中文)6-7 在一個數組中實現兩個堆疊

6-7 在一個數組中實現兩個堆疊(20 分)

本題要求在一個數組中實現兩個堆疊。

函式介面定義:

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

其中Tag是堆疊編號,取1或2;MaxSize堆疊陣列的規模;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:如果堆疊已滿,Push函式必須輸出“Stack Full”並且返回false;如果某堆疊是空的,則Pop函式必須輸出“Stack Tag Empty”(其中Tag是該堆疊的編號),並且返回ERROR。

裁判測試程式樣例:

#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;
}

/* 你的程式碼將被嵌在這裡 */

輸入樣例:

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

輸出樣例:

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
作者: 陳越單位: 浙江大學時間限制: 400ms記憶體限制: 64MB程式碼長度限制: 16KB
Stack CreateStack( int MaxSize ){
  Stack stack = (Stack)malloc(sizeof(Stack));
  stack->Data = (ElementType*)malloc(MaxSize*sizeof(ElementType));
  stack->Top1 = -1;
  stack->Top2 = MaxSize;
  stack->MaxSize = MaxSize;
  return stack;
}

bool Push( Stack S, ElementType X, int Tag ){
  if(S->Top1+1==S->Top2){
    printf("Stack Full\n");
    return false;
  }
  if(Tag == 1){
    S->Data[++(S->Top1)] = X;
  }
  if(Tag == 2){
    S->Data[--(S->Top2)] = X;
  }
  return true;
}

ElementType Pop( Stack S, int Tag ){
  if(Tag == 1 && S->Top1 == -1 || Tag == 2 && S->Top2 == S->MaxSize){
    printf("Stack %d Empty\n",Tag);
    return ERROR;
  }
  return Tag==1?S->Data[(S->Top1)--]:S->Data[(S->Top2)++];
}