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

思路：准备2个栈，一个保存入栈序列，一个保存出栈序列。将1-n入栈，在入栈过程中如果入栈序列和出栈序列一致，就出使栈顶元素出栈，出栈后还是相等就继续出栈（此处一个循环，直到栈顶元素不等位置），然后继续入栈

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

int main()
{
    int m,n,k;//栈容量，序列长度，要检查的序列数
    scanf("%d %d %d",&m,&n,&k);

    while(k--)//k个要检查的序列
    {
        bool flag = 1;//一开始假设这个序列没错
        stack<int> st;//入栈序列
        while(!st.empty())//入栈之前栈必须要空
            st.pop();

        int arr[n+1];//保存出栈的序列
        int cur =1;//栈顶指针
        for(int i=1;i<=n;i++)//输入出栈序列
        {
            scanf("%d",&arr[i]);
        }

        for(int i=1;i<=n;i++)//入栈序列开始进栈
        {
            st.push(i);
            if(st.size()>m)//超过了栈的容量就停止
            {
                flag = false;
                
            }

            while(!st.empty()&& st.top()==arr[cur])//栈顶元素相同就出栈
            {
                st.pop();
                cur++;

            }

        }

        if(st.empty()&&flag == true)
            cout<<"YES\n";
        else
            cout<<"NO\n";


    }
    return 0;
}