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  • 1051 Pop Sequence 入栈 出栈模拟

    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<cstring>
    #include<algorithm>
    #include<stack>
    using namespace std;
    const int inf=1005;
    int main()
    {
    int n,m,t,a[inf];
    stack<int> s;
    cin>>n>>m>>t;
    while(t--)
    {
    for(int i=0;i<m;i++)
    cin>>a[i];
    int i=1,j=0;
    while(j<m&&i<=m)
    {
    s.push(i);
    if(s.size()>n)
    break;
    //cout<<s.size()<<endl;
    while(s.top()==a[j]&&!s.empty()&&j<m&&i<=m)
    {
    //cout<<s.top()<<"%%% "<<endl;
    j++;
    s.pop();
    //cout<<s.size()<<endl;
    if(s.empty())
    {
    //cout<<s.top()<<endl; 当栈为空时不能使用s.top(),程序会中断,坑了我半个下午mmp
    break;
    }
    //cout<<s.top()<<"%%% "<<endl;
    }
    i++;
    }
    if(!s.empty())
    cout<<"NO"<<endl;
    else
    cout<<"YES"<<endl;
    while(!s.empty())
    s.pop();
    }
    return 0;
    }

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  • 原文地址:https://www.cnblogs.com/liuzhaojun/p/11195716.html
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