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
阿新 • • 發佈:2018-12-24
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: 16KBStack 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)++]; }