pat1051

1051. Pop Sequence (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

题目要求：对于1到n的输入，按照不同的方式入栈出栈，栈的大小为m，对应的出栈结果是否可以为列出的结果；

解题思路，设计一个stack装1—n和一个queue装对应的结果，每次放一个入栈和入队，当栈顶和队首元素相同的时候就出栈和出队，直到其中一个为空。最后判断是否两个都为空，并且没有超出栈的大小。

#include
#include
#include
using namespace std;
int main(){
    int m,n,k;
    cin>>m>>n>>k;
    for(int i=0;i s;
        queue q;
        for(int i=0;i>tmp;
            s.push(tmp1++);
            q.push(tmp);
            if(s.size()>m)
                flag=1;
            while(!s.empty()&&!q.empty()&&s.top()==q.front()){
                s.pop();
                q.pop();
            }
        }
        if(!flag&&s.empty()&&q.empty())
            cout<<"YES\n";
        else cout<<"NO\n";
    }
    return 0;
}