【PAT甲】1051 Pop Sequence (25分)判断出栈顺序的合法性

problem

1051 Pop Sequence (25分)
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

• 给定一个大小为n的栈，按照1,2,3,…n的顺序依次入栈
• 给出k个长为m的出栈序列，判断是否合法

solution

直观的思路就是将入栈序列一个一个入栈，与出栈序列相比较，一样就出栈，不一样就继续入栈，当入栈序列和出栈序列都为空时，表示出栈顺序合法。

建立一个辅助栈 把输入的第一个序列中的数字一个一个压入该辅助栈，并按照第二个序列的顺序从该栈中弹出数字。
(1)如果元素是栈顶的元素，则pop出来；
(2)如果不是栈顶元素，则根据入栈顺序将没入栈的元素push进栈，直到将该元素push栈中，然后将该元素pop出来；如果push完所有元素都没有找到该元素，那么这个出栈顺序是错误的。最后辅助栈为空栈，并且遍历完了出栈序列的最后一个元素（二者缺一不可），那么该序列就是 一个弹出序列

#include<iostream>
#include<stack>
using namespace std;
int a[1010];

int main(){
	int n, m, k;
	cin>>n>>m>>k;
	while(k--){
		for(int i = 1; i <= m; i++)cin>>a[i];
		int ok = 1, cur = 0;
		stack<int>s;
		for(int i = 1; i <= m; i++){
			if(s.empty())s.push(++cur);
			while(s.top()!=a[i]&&s.size()<=n)s.push(++cur);
			if(s.size()>n){
				ok = 0; break;
			}
			if(s.top()==a[i]){
				s.pop(); continue;
			}else{
				ok = 0; break;
			}
		}
		if(ok==1&&s.size()==0){
			cout<<"YES\n";
		}else{
			cout<<"NO\n";
		}
	}
	return 0;
}


  

 

  
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文章来源: gwj1314.blog.csdn.net，作者：小哈里，版权归原作者所有，如需转载，请联系作者。

原文链接：gwj1314.blog.csdn.net/article/details/107982111