1051 Pop Sequence

1051 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

直接使用STL模拟一下即可

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <stack>

using namespace std;
const int maxn = 1005;
int n,m,k,a[maxn];
stack<int> s;
int main()
{
    scanf("%d%d%d",&n,&m,&k);
    while(k--)
    {
        for(int i = 0; i < m;i ++)
        {
            scanf("%d",&a[i]);
        }
        while(!s.empty())
            s.pop();
        int num = 0,index = 0;
        for(int i = 1; i <= m;i ++)
        {
            while(!s.empty() && s.top() == a[index])
                s.pop(),index++;
            if(s.size()+1 > n) break;
            s.push(i);
        }
        while(!s.empty() && s.top() == a[index])
            s.pop(),index++;
        if(index == m )
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}