PAT_A 1051. Pop Sequence (25)

1051. Pop Sequence (25)

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 pushsequence), and K (the number of pop sequences to be checked). Then K lines follow, eachcontains 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 ofthe stack, or "NO" if not.Sample Input:5 7 51 2 3 4 5 6 73 2 1 7 5 6 47 6 5 4 3 2 15 6 4 3 7 2 11 7 6 5 4 3 2Sample Output:YESNONOYESNO

• 分析：

  • 题目：检查所给序列是否是正确的栈push pop序列。
  • 解题：这个我们通过观察，可以发现对于每次的每次pop之后如果发生push就会形成一个明显的片段序列。规律是这样的:1,不发生push的pop是非增序列 ai,ai+1,ai+2,…an;2,一但发生push(bk)，在pop就会形成比前一个序列大的元素 bk,并且bk 之间是增长关系。如5 6 4 3 7 2 1,可以看出在5 6 7 之前分别进行了push(push 1 2 3 4 5 pop 5,push 6 ,pop 6 4 3,push 7,pop 7 2 1),

• code:

#include<iostream>#include<cstdio>using namespace std;int seq[1010];int main(){    freopen("in","r",stdin);    int M,N,K,tmp,count,pre,split,isRight;    scanf("%d%d%d",&M,&N,&K);    for(int i=0;i<K;i++)    {        pre=split=0;        isRight=1;        for(int j=0;j<N;j++)        {          scanf("%d",&tmp);          if(tmp>pre)//新的划分          {            if(tmp<split)//划分值非递增,非法                isRight=0;            split=tmp;            count=1;          }else            count++;          if(count>M)//超过栈的容量            isRight=0;          pre=tmp;        }        if(isRight==0)          printf("NO\n");        else          printf("YES\n");    }    return 0;}

• AC