【PAT】1051. Pop Sequence (25)

题目链接：http://pat.zju.edu.cn/contests/pat-a-practise/1051

题目描述：

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：

考察栈的操作。

输入的序号是1,2,3......N。

以

5 6 4 3 7 2 1

为例，要pop 5，就必须先push 1, push 2, push 3, push 4, push5, 此时栈顶元素为5，刚好匹配，所以进行pop操作。

设置一个变量temp，由1开始自增。当栈顶元素与输入的元素不匹配且栈元素个数未满M时就进行入栈操作：sta.push(temp++);。

如果栈中的元素超过了M值，则说明出现了错误，这种出栈方式是不对的。

参考：

http://blog.dayandcarrot.net/2013/05/1051-pop-sequence-25/

http://tech-wonderland.net/blog/pat-1051-pop-sequence.html

http://blog.csdn.net/sunbaigui/article/details/8656793

参考代码：

#include<iostream>
#include<stack>
using namespace std;

int _tmain(int argc, _TCHAR* argv[])
{
	int M; //maximum capacity of the stack
	int N; //the length of push sequence
	int K; //the number of pop sequence to be checked
	cin>>M>>N>>K;
	int i,j;
	int input,temp;
	bool flag = true;
	stack<int> sta;
    
	for(i=0; i<K; i++)
	{
		temp = 1;
		flag = true;
		for(j=0; j<N; j++)
		{
			cin>>input;
			while( sta.size() <= M && flag )
			{
				if(sta.empty() || sta.top() != input)
				{
					sta.push(temp ++);
				}
				else if(sta.top() == input)
				{
					sta.pop();
					break;
				}			
			}
			if(sta.size() > M)
			{flag = false;}
		}

		if(flag)
			cout<<"YES"<<endl;
		else 
			cout<<"NO"<<endl;
		while(!sta.empty())
			sta.pop();
	}
	return 0;
}