【PTA-线性结构4】Pop Sequence

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

思路:

模拟堆栈操作验证出栈顺序是否合法。先用数组储存出栈顺序，再进行堆栈操作。如果要出栈的元素大于要入栈的元素，则循环入栈，直到栈顶元素等于该元素，则将弹出栈顶。如果要出栈的元素等于栈顶元素，则直接弹栈。如果要出栈的元素小于栈顶元素，则认为序列不合法，因为入栈元素是从小到大递增的，小的元素被压在下面出不来。如果循环过程都正常且循环结束后栈为空，则认为该出栈顺序合法。

代码:

#include <iostream>
#include <stdio.h>
#include <stdlib.h>

int m;
typedef struct node
{
    int data;
    node *next;
} Node, *Stack;

Stack createStack()
{
    Stack stack = new Node;
    stack->next = NULL;
    stack->data = 0;
    return stack;
}

int push(Stack stack, int data)
{
    if (stack->data == m)
        return 0;
    Stack node = new Node;
    node->data = data;
    node->next = stack->next;
    stack->next = node;
    stack->data++;
    return 1;
}

int pop(Stack stack)
{
    int data;
    Stack popNode = stack->next;
    data = popNode->data;
    stack->next = popNode->next;
    delete popNode;
    stack->data--;
    return data;
}

int main()
{
    int n, num, isTrue, top;
    std::cin >> m >> n >> num;
    int a[n];
    for (int i = 0; i < num; ++i)
    {
        Stack stack = createStack();
        isTrue = 1;
        for (int j = 0; j < n; ++j)
        {
            std::cin >> a[j];
        }
        for (int j = 0, k = 1; j < n && isTrue; ++j)
        {
            while (k <= a[j])
            {
                if (push(stack, k++) == 0)
                {
                    isTrue = 0;
                    break;
                }
            }
            top = stack->next->data;
            if (top == a[j])
                pop(stack);
            else
            {
                isTrue = 0;
                break;
            }
        }
        if (isTrue && stack->data == 0)
            std::cout << "YES\n";
        else
            std::cout << "NO\n";
    }
    return 0;
}