PAT A1051 Pop Sequence弹出序列 [栈]

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.

给你一个最多能存放M个数的栈。把N个数以1,2,3，。。。N的顺序放入栈，并随机弹出。你需要判断一个数字序列是否有可能是栈的弹出序列。比如说，如果M是5，N是7，我们能从栈得到序列1,2,3,4,5,6,7 而不是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.

第一行包含3个数字（每一个都不会大于1000）：M N K(需要检测的弹出序列个数)。然后会有K行，每一行都是包含N个数字的弹出序列。数字之间用空格隔开

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.

对于每一个弹出序列，如果它确实是弹出序列则输出"YES"，否则输出"NO"。

思路：模拟入栈，让1-N依次入栈，入栈过程中如果入栈元素恰好等于出栈序列中待出栈元素，就让栈顶元素出栈。同时把出栈元素指针后移一个。

#include<cstdio>
#include<stack>
using namespace std;
const int maxn=1010;

int arr[maxn];//保留题目给定的出栈序列
stack<int> st;//定义栈st,用以存放int型元素

int main(){
	int m,n,T;
	scanf("%d %d %d",&m,&n,&T);

	while(T--){  //循环T次
		//1.清空栈 
		while(!st.empty()){
			st.pop();
		}
        //2.接收出栈序列
		for(int i=1;i<=n;i++){
			scanf("%d",&arr[i]);
		}
		int current = 1;//指向出栈序列中的待出栈元素
		bool flag=true;
        //3.模拟入栈出栈
		for(int i=1;i<=n;i++){
			st.push(i);//i入栈
			if(st.size()>m){
				flag=false;
				break;
			} 
			
			while(!st.empty()&&st.top()==arr[current]){
				st.pop();
				current++;
			}
		} 
		//4.输出
		if(st.empty()==true&&flag==true){
			printf("YES\n");
		}else{
			printf("NO\n");
		}
	}
	return 0;
}