数据结构练习 02-线性结构3. Pop Sequence (25)

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

#include<iostream>
#include<stack>
using namespace std;int main()
{
    int M; //maximum capacity of the stack
    int N; //the length of push sequence
    int K; //the number of pop sequence to be checked
    cin >> M >> N >> K;
    int i, j;
    int input, temp;
    bool flag = true;
    stack<int> sta;    for (i = ; i < K; i++)
    {
        temp = ;
        flag = true;
        for (j = ; j < N; j++)
        {
            cin >> input;
            while (sta.size() <= M && flag)
            {
                if (sta.empty() || sta.top() != input)
                {
                    sta.push(temp++);
                }
                else if (sta.top() == input)
                {
                    sta.pop();
                    break;
                }
            }
            if (sta.size() > M)
            {
                flag = false;
            }
        }        if (flag)
            cout << "YES" << endl;
        else
            cout << "NO" << endl;
        while (!sta.empty())
            sta.pop();
    }
    return ;
}