1051 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

题意

给定一个最多能存 M 个数字的栈，将 1∼N 按顺序压入栈中，过程中可随机弹出栈顶元素。

当 N 个数字都经历过入栈和出栈后，我们按照元素出栈的顺序，可以得到一个弹出序列。

现在给定一系列 1∼N 的随机排列序列，请你判断哪些序列可能是该栈的弹出序列。

例如，当 N=7,M=5 时，1, 2, 3, 4, 5, 6, 7可能是该栈的弹出序列，而 3, 2, 1, 7, 5, 6, 4 不可能是该栈的弹出序列。

思路

这题是一道栈模拟的题目，每次模拟操作只有两个选择：

1.推入元素，即如果当前栈顶元素不等于当前需要输出的元素，则推入新的元素。

2.弹出元素，即如果当前栈顶元素等于当前需要输出的元素，则弹出栈顶元素，直至不相等为止。

而判断该弹出序列是否满足要求同样有两个依据：

1.栈中元素不能大于 m 。

2.做完所有操作后，栈中不能有元素即所有栈顶元。

代码

#include<bits/stdc++.h>
using namespace std;
const int N = 1010;
int a[N];
int m, n, k;
bool check()
{
    stack<int> stk;
    for (int i = 1, j = 0; i <= n; i++)
    {
        stk.push(i);
        //已经超出栈的容量，则返回false
        if (stk.size() > m)    return false;
        //判断当前输出的元素是否与栈顶元素相等
        while (stk.size() && stk.top() == a[j])
        {
            stk.pop();
            j++;
        }
    }
    return stk.empty();
}
int main()
{
    cin >> m >> n >> k;
    while (k--)
    {
        for (int i = 0; i < n; i++)    cin >> a[i];
        if (check()) puts("YES");
        else    puts("NO");
    }
    return 0;
}