PAT Advanced 1051 Pop Sequence (25) [栈模拟]

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

题目分析

已知一个栈限制容量最大为M，push操作压入1,2,3...N，已知K个序列，判断序列是否可能是出栈顺序

解题思路

1. 1~N依次入栈，在入栈过程中，由p指针指向序列中当前等待出栈的元素
2. 如果栈顶元素恰好等于序列当前等待出栈的元素，让栈顶元素出栈，p指针后移
3. 继续循环2的操作，知道栈顶元素不等于序列中当前等待出栈的元素
4. 继续1~N的压入
5. 判断是否为出栈顺序
   • 如果压栈过程中栈大小>=M，说明栈满，该序列不是出栈顺序，退出；
   • 如果压栈执行结束后，栈中有剩余元素或者p指针未指向序列末尾，证明该序列不是出栈顺序

Code

Code 01

#include <iostream>
#include <stack>
using namespace std;
int main(int argc,char * argv[]) {
	int n,m,k;
	scanf("%d%d%d",&n,&m,&k);
	int seq[m]= {0};
	for(int i=0; i<k; i++) {
		for(int j=0; j<m; j++) {
			scanf("%d",&seq[j]);
		}
		stack<int> sk;
		int j,p = 0;
		for(j=1; j<=m; j++) {
			if(sk.size()>=n)break;
			sk.push(j);
			while(!sk.empty()&&sk.top()==seq[p]) {
				sk.pop();
				p++;
			}

		}
		if(j<=n||!sk.empty())printf("NO
");
		else printf("YES
");
	}
	return 0;
}