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 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

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;
}