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 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

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