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1051. Pop Sequence (25)
阅读量:2148 次
发布时间:2019-04-30

本文共 1497 字,大约阅读时间需要 4 分钟。

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
 
#include 
#include
#include
#include
#include
using namespace std; int main() { int n, m, k; int i, j; int in, out; cin>>n>>m>>k; while(k--) { stack
q; int flag = 0; in = 1; for(i = 0; i < m; i++) { cin>>out; while(q.empty() || out!=q.top()) { q.push(in); if(q.size()>n) { flag = 1; break; } in++; } if(!flag && !q.empty() && q.top()==out) q.pop(); } if(flag == 0) cout<<"YES"<

 

 

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