题目

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

分析

题目大意就是一个容量限制为 m 的栈，一组数 1,2,3,…,n 顺序入栈，随机出栈，问出栈序列是否可能
刚开始想的是找出一组规律，根据规律判断该出栈顺序是否合理，找着找着规律好像不太好找，索性进行模拟
使用 v 数组存储输入序列，定义栈 s，让 1,2,…n 顺序入栈，定义 cur 指示当前 v 数组下标， 当 v[cur] 等于栈顶元素（s.top()）且栈 s 不为空，不断出栈且 cur+1 直到不等为止，当 1…n 入栈结束，或者栈 s 已满，退出入栈循环，最后判断 cur 与 n 的大小关系决定该顺序是否可能
比如例序列 5 6 4 3 7 2 1，此时 m = 5，即栈的最大容量为 5

当前入栈元素 栈 s v 数组 cur v[cur] 出栈
1 1 5 6 4 3 7 2 1 1 5 无
2 1 2 5 6 4 3 7 2 1 1 5 无
3 1 2 3 5 6 4 3 7 2 1 1 5 无
4 1 2 3 4 5 6 4 3 7 2 1 1 5 无
5 1 2 3 4 5 5 6 4 3 7 2 1 1 5 5
6 1 2 3 4 6 5 6 4 3 7 2 1 2 6 6
无 1 2 3 4 5 6 4 3 7 2 1 3 4 4
无 1 2 3 5 6 4 3 7 2 1 4 3 3
7 1 2 7 5 6 4 3 7 2 1 5 7 7
2 1 2 5 6 4 3 7 2 1 6 2 2
1 1 5 6 4 3 7 2 1 7 1 1 
#include<iostream>
#include<vector>
#include<stack>
using namespace std;
int main(){
	int m,n,k;
	cin>>m>>n>>k;
	for(int i=0;i<k;i++){
		stack<int> s;
		vector<int> v(n + 1);
		for(int j=1;j<=n;j++)
			cin>>v[j];
		int cur=1;
		for(int j=1;j<=n;j++){
			s.push(j);
			if(s.size() > m)
				break;
			while(!s.empty() && s.top()==v[cur]){
				s.pop();
				cur++;
			}
		}
		if(cur == n+1)
			cout<<"YES"<<endl;
		else
			cout<<"NO"<<endl;
	}
	return 0;
}