02-线性结构4 Pop Sequence (25分)

02-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

提交代码OC通过:

提交代码:

#include <stdio.h>
#include <stdlib.h>
#define ERROR -1

typedef int ElemType;
typedef int Position;
struct SNode {
    ElemType *Data;
    Position Top;
    int MaxSize;
};

typedef struct SNode *Stack;

Stack CreateStack(int MaxSize) {
    Stack S = (Stack)malloc(sizeof(struct SNode));
    S->Data = (ElemType*)malloc(MaxSize * sizeof(ElemType));
    S->Top = -1;
    S->MaxSize = MaxSize;
    return S;
}

void DestroyStack(Stack S) {
    if (S == NULL) {
        return;
    }
    if (S->Data) {
        free(S->Data);
    }
    free(S);
}

int IsFull(Stack S)
{
    return (S->Top == S->MaxSize - 1);
}

int Push(Stack S, ElemType X) {
    if (IsFull(S)) {
        return 0;
    }
    else {
        S->Data[++(S->Top)] = X;
        return 1;
    }
}

int IsEmpty(Stack S) {
    return (S->Top == -1);
}

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

int main() {
    int maxStackSize, seqSize, stackNum;
    scanf("%d %d %d", &maxStackSize, &seqSize, &stackNum);
    int* outputFlag = (ElemType*)malloc(sizeof(ElemType) * stackNum);
    for (int i = 0; i < stackNum; ++i) {
        outputFlag[i] = 0;
    }
    ElemType* sequence = (ElemType*)malloc(sizeof(ElemType)*seqSize);
    Position pOrigin = 0;
    Position pSequence = 0;
    for (int i = 0; i < stackNum; ++i) {//对处理stackNum个Sequence的循环
        for (int j = 0; j < seqSize; ++j) {
            scanf("%d", sequence + j);
        }
        pSequence = 0;
        Stack S = CreateStack(maxStackSize);
        ElemType count = 1;
        while (pSequence < seqSize) {
            if (S->Top == -1 || S->Data[S->Top] < sequence[pSequence]) {
                if (!Push(S, count++)) {
                    break;
                }
            }
            else if (S->Data[S->Top] == sequence[pSequence]) {
                Pop(S);
                ++pSequence;
            }
            else {
                break;
            }
        }
        if (!IsEmpty(S) || pSequence != seqSize) {
            outputFlag[i] = 0;
            //printf("No
");
        }
        else {
            outputFlag[i] = 1;
            //printf("Yes
");
        }
        DestroyStack(S);
    }
    for (int i = 0; i < stackNum; ++i) {
        if (outputFlag[i] == 0) {
            printf("NO
");
        }
        else {
            printf("YES
");
        }
    }
    return 0;
}
原文地址:https://www.cnblogs.com/2018shawn/p/13251903.html