• 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;
    }
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  • 原文地址:https://www.cnblogs.com/2018shawn/p/13251903.html
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