数据结构练习 02-线性结构3. Pop Sequence (25)

时间:2023-03-09 01:35:45
数据结构练习 02-线性结构3. 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
#include<iostream>
#include<stack>
using namespace std; int main()
{
int M; //maximum capacity of the stack
int N; //the length of push sequence
int K; //the number of pop sequence to be checked
cin >> M >> N >> K;
int i, j;
int input, temp;
bool flag = true;
stack<int> sta; for (i = ; i < K; i++)
{
temp = ;
flag = true;
for (j = ; j < N; j++)
{
cin >> input;
while (sta.size() <= M && flag)
{
if (sta.empty() || sta.top() != input)
{
sta.push(temp++);
}
else if (sta.top() == input)
{
sta.pop();
break;
}
}
if (sta.size() > M)
{
flag = false;
}
} if (flag)
cout << "YES" << endl;
else
cout << "NO" << endl;
while (!sta.empty())
sta.pop();
}
return ;
}

数据结构练习 02-线性结构3. Pop Sequence (25)