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PAT-A-1051. Pop Sequence (25)

2017-05-13 23:13 387 查看


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

#include<iostream>
#include<cstdio>
#include<stack>

using namespace s
4000
td;

const int maxn = 1010;
int arr[maxn];
stack<int> st;
int main()
{
int m, n, t;
cin >> m >> n >> t;
while (t--)
{
while (!st.empty())
st.pop();
for (int i = 1; i <= n; i++)
cin >> arr[i];

int current = 1;
int flag = 1;
for (int i = 1; i <= n; i++)
{
st.push(i);
if (st.size() > m)
{
flag = 0;
break;
}

while (!st.empty() && st.top() == arr[current])
{
st.pop();
current++;
}
}
if (st.empty() == true && flag == 1)
cout << "YES" << endl;
else
cout << "NO" << endl;
}
system("pause");
return 0;
}
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