PAT 1051. Pop Sequence (25) 栈

1051. Pop Sequence (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 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 2

Sample Output:

YESNONOYESNO

用数组模拟了栈练练手。当然用stack更方便。

#include<string>#include<stdio.h>#include<iostream>#include<string.h>#include<queue>#include<algorithm>#include<map>#include<vector>using namespace std;int a[1005];int b[1005];  int main(){   int m,n,k;   int p=0;   cin>>m>>n>>k;   for(int i=0;i<k;i++)   {   p=0;      int flag=0;      int cnt=0;      a[0]=1;      for(int j=0;j<n;j++)           cin>>b[j];       for(int j=0;j<n;j++)        {             if(cnt<b[j])             {                while(cnt<b[j])                {                     a[p++]=cnt+1;                     cnt++;                }                if(p>m)                {                    flag=1;                    break;                }                p--;             }              else              {                  while(a[p-1]!=b[j]&&p>=1)                      p--;                      p--;                  if(p<0)                  {                      flag=1;                      break;                  }              }        }       if(flag==1) cout<<"NO"<<endl;       else cout<<"YES"<<endl;     }    return 0;}