PAT甲级1051 栈的模拟

最新推荐文章于 2022-12-02 22:14:11 发布

原创最新推荐文章于 2022-12-02 22:14:11 发布 · 465 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#PAT

刷子刷题专栏收录该内容

30 篇文章

订阅专栏

本文提供了一种判断给定序列是否为有限容量栈可能弹出序列的有效算法实现。通过实例演示了如何利用栈的数据结构特性进行验证。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目

1051 Pop Sequence (25)（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

题解：

#include<iostream>
#include<vector>//前两位不能有6，7,7不能在5，6之前
#include<stack>
using namespace std;
int main(){
  int m,n,k;
  cin>>m>>n>>k;


  for(int i=0;i<k;i++){
    bool flag=false;
    stack<int> s;
    vector<int> v(n);//0~size-1
    for(int j=0;j<n;j++){
    scanf("%d",&v[j]);
  }
    int index=0;
    for(int j=1;j<=n;j++){
      s.push(j);
      if(s.size()>m) break;
      while(!s.empty()&&s.top()==v[index]){
        s.pop();
        index++;
      }
    }
    if(index==n) flag=true;
    if(flag) printf("YES\n");
    else printf("NO\n");

  }
  return 0;
}