【线性结构】A1051 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：

• 定义动态数组v存储待判的序列，定义栈s模拟入栈出栈操作；
• n个数依次进栈，如果栈内个数超过m则break，否则当栈非空时判断栈顶元素和序列中对应位置元素是否相等，相等则弹出否则入栈；
• 最后判断序列是否已经遍历完，如果没有就是NO，否则就是YES

#include <iostream>
#include <stack>
#include <vector>
using namespace std;
int main()
{
    int m, n, k;
    cin >> m >> n >> k;
    for (int i = 0; i < k; i++) {
        stack<int>s;
        vector<int>v(n);
        for (int j = 0; j < n; j++) {
            cin >> v[j];
        }
        int cur=0;//v中元素的下标
        for (int j = 1; j <= n; j++) {
            s.push(j);
            if (s.size() > m)break;
            while (!s.empty() && s.top() == v[cur]) {
                s.pop();
                cur++;
            }
        }
        if (cur < n)
            cout << "NO" << endl;
        else
            cout << "YES" << endl;
    }
    return 0;
}