Given a stack which can keep numbers at most. Push numbers in the order of 1, 2, 3, …, 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 is 5 and 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): (the maximum capacity of the stack), (the length of push sequence), and (the number of pop sequences to be checked). Then lines follow, each contains a pop sequence of 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
题意
有一个容量限制为M的栈,先分别把1,2,3,…,n依次入栈,在入栈的过程中可以随机出栈。给出一系列出栈顺序,问这些出栈顺序是否可能。
思路
模拟。
代码
#include <iostream>
#include <stack>
#include <vector>
using namespace std;
int main() {
cin.tie(nullptr);
cout.tie(nullptr);
ios::sync_with_stdio(false);
int m, n, k;
cin >> m >> n >> k;
for (int i = 0; i < k; ++i) {
stack<int> stk;
vector<int> target(n);
for (int j = 0; j < n; ++j)
cin >> target[j];
for (int j = 1, pos = 0; j <= n; ++j) {
if (stk.size() == m)
break;
stk.push(j);
while (not stk.empty() && stk.top() == target[pos]) {
stk.pop();
++pos;
}
}
if (stk.empty())
cout << "YES\n";
else
cout << "NO\n";
}
}