线性结构4 Pop Sequence (25 分)

Given a stack which can keep M numbers at most. Push N numbers in the order of 1, 2, 3, …, N and pop randomly. You are supposed to tell if a given sequence of numbers is a possible pop sequence of the stack. For example, if M is 5 and N is 7, we can obtain 1, 2, 3, 4, 5, 6, 7 from the stack, but not 3, 2, 1, 7, 5, 6, 4.

Input Specification:

Each input file contains one test case. For each case, the first line contains 3 numbers (all no more than 1000): M (the maximum capacity of the stack), N (the length of push sequence), and K (the number of pop sequences to be checked). Then K lines follow, each contains a pop sequence of N numbers. All the numbers in a line are separated by a space.

Output Specification:

For each pop sequence, print in one line “YES” if it is indeed a possible pop sequence of the stack, or “NO” if not.

Sample Input:

5 7 5
1 2 3 4 5 6 7
3 2 1 7 5 6 4
7 6 5 4 3 2 1
5 6 4 3 7 2 1
1 7 6 5 4 3 2

Sample Output:

YES
NO
NO
YES
NO

Code

#include 
#include 
#include 
#define MAXSIZE 1000
#define ERROR -1

typedef int ElementType;
typedef int Position;
typedef struct SNode *Stack;
struct SNode{
    ElementType *Data;
    Position Top;
    int Size;
};

Stack InitStack(int Size);
bool IsFull(Stack PtrS);
bool IsEmpty(Stack PtrS);
bool Push(Stack PtrS,ElementType X);
ElementType Pop(Stack PtrS);
void ClearStack(Stack PtrS);
/*算法：（此题用顺序结构表示栈比较简单）
1.从键盘中将需要checked的序列存储到数组num中；
2.从第一个数开始判断，第一个数字需要出栈,必须是1，2，3...的顺序进栈直到num[0];
若栈满还未遍历到num[0]，输出“NO”;否则继续判断第二个数
3.若num[h]>num[h-1]，則继续进栈，直到遍历到，出栈，继续判断下一个数，
若栈满还未遍历到num[0]，输出“NO”；
  若num[h]
int main()
{
    int m,n,k;
    scanf("%d %d %d",&m,&n,&k);
    Stack S = InitStack(m);
    int num[n];
    for(int i=0;i<k;i++)
    {
        int cnt = 0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&num[i]);
        }

        while(cnt!=num[0])
        {
            if(!Push(S,++cnt))
            {
                printf("NO\n");
                goto out;
            }
//            printf("S->Data[%d] = %d\n",S->Top,S->Data[S->Top]);
        }
        Pop(S);


        for(int h=1;h<n;h++)
        {
            if(num[h]>num[h-1])
            {
                while(cnt!=num[h])
                    {
                        if(!Push(S,++cnt))
                        {
                            printf("NO\n");
                            goto out;
                        }
//                            printf("S->Data[%d] = %d\n",S->Top,S->Data[S->Top]);

                    }
                    Pop(S);
            }else
            {
                if(Pop(S)!=num[h])
                {
                    printf("NO\n");
                    goto out;
                }
            }
        }
        printf("YES\n");
        out:
            ClearStack(S);
            continue;
    }
    return 0;
}

Stack InitStack(int Size)
{
    Stack S = (Stack)malloc(sizeof(struct SNode));
    S->Data = (ElementType *)malloc(Size*sizeof(ElementType));
    S->Top = -1;
    S->Size = Size;
    return S;
}

bool IsFull(Stack PtrS)
{
    return PtrS->Top == PtrS->Size-1;
}

bool IsEmpty(Stack PtrS)
{
    return PtrS->Top == -1;
}

bool Push(Stack PtrS,ElementType X)
{
    if(IsFull(PtrS))
    {
        return false;
    }else
    {
        PtrS->Top++;
        PtrS->Data[PtrS->Top] = X;
        return true;
    }
}

ElementType Pop(Stack PtrS)
{
    if(IsEmpty(PtrS))
    {
        return ERROR;
    }else
    {
        return PtrS->Data[(PtrS->Top)--];
    }
}

void ClearStack(Stack PtrS)
{
    PtrS->Top = -1;
}