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

#include<iostream>
#include<stack>
using namespace std;

const int maxn = ;

int check(int *a,int n,int m);

int main()
{
    int arr[maxn] = {};
    int n,m,k;
    scanf("%d%d%d",&m,&n,&k);

    for (int i = ; i < k; i++)
    {
        for (int j = ; j <= n; j++)
        {
            scanf("%d",&arr[j]);
        }
        if(check(arr,n,m))
        {
            printf("YES\n");
        }
        else
        {
            printf("NO\n");
        }
    }
    return ;
}

int check(int *a,int n,int m)
{
    int iRet = -;
    stack<int> s;
    int current = ;
    bool flag = true;
    for (int i = ; i <= n; i++)
    {
        s.push(i);
        if (s.size() > m)
        {
            flag = false;
            break;
        }
        while (!s.empty() && a[current] == s.top())
        {
            current++;
            s.pop();
        }
    }

    if (flag && s.empty())
    {
        iRet = ;
    }
    else
    {
        iRet = ;
    }

    return iRet;
}