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

提交代码

#include<fstream>
#include <cstdio>
#include <iostream>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<string>
#include<cmath>
#include<stack>
using namespace std;

int main()
{

int n,m,k;
scanf("%d%d%d",&m,&n,&k);

for(int i=0;i<k;i++)
{
vector<int> v(n+1);
for(int j=1;j<=n;j++)
cin>>v[j];
stack<int> s;
int current=1;
for(int j=1;j<=n;j++)
{
s.push(j);
if(s.size()>m) break;
while(!s.empty()&&s.top()==v[current])
{
s.pop();
current++;
}

}
if(current==n+1) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}

#ifdef _DEBUG
system("pause");
#endif

return 0;

}