- 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
/*
PAT 甲级 1051 栈模拟
*/
#include <iostream>
#include <vector>
#include <stack>
using namespace std;
int main() {
int m, n, k;
scanf("%d%d%d", &m, &n, &k);
for(int i = 0; i< k; i++){
vector<int> v(n);
stack<int> s;
for(int j = 0; j < n; j++){
scanf("%d", &v[j]);
}
int index = 0;
for(int j = 0; j < n; j++){
s.push(j+1); //这里位置不能交换
if(s.size() > m) break;
while(!s.empty() && s.top() == v[index]){
s.pop();
index++;
}
}
if(index == n) cout << "YES" << endl;
else cout << "NO" << endl;
}
return 0;
}
