1051 Pop Sequence (25 分)-优快云博客

本文链接：https://blog.youkuaiyun.com/while_BLUE_/article/details/116379666

本文介绍了一种算法，用于判断给定的数列是否能由一个具有特定最大容量的栈通过压栈和弹栈操作得到。该算法适用于解决计算机科学中的数据结构问题，特别是与栈操作相关的问题。

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原题

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

常规栈题，有限栈。
首先将序列存储进数组，然后再处理序列问题：在内循环中将1-n依次入栈，并在入栈时检测栈顶元素是否等于数组元素，若相同则出栈。最后通过检测每轮下来的栈是否为空以及元素是否超过栈容量来判断Y/N。

代码

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int m,n,k;
    cin>>m>>n>>k;
    stack<int> st;
    int a[1000];
    for(int i=0;i<k;i++)
    {
        int index=1;
        bool flag=true;
        while(!st.empty())st.pop();
        for(int j=1;j<=n;j++)
        {
            cin>>a[j];
        }
        for(int j=1;j<=n;j++)
        {
            st.push(j);
            if(st.size()>m){
                flag=false;
                break;
            }
            while(!st.empty()&&st.top()==a[index])
            {
                st.pop();
                index++;
            }
        }
        if(st.empty()&&flag)cout<<"YES\n";
        else cout<<"NO\n";
    }
    return 0;
}