Hdu 1022

题目描述：

As the new term comes, the Ignatius Train Station is very busy nowadays. A lot of student want to get back to school by train(because the trains in the Ignatius Train Station is the fastest all over the world ^v^). But here comes a problem, there is only one railway where all the trains stop. So all the trains come in from one side and get out from the other side. For this problem, if train A gets into the railway first, and then train B gets into the railway before train A leaves, train A can't leave until train B leaves. The pictures below figure out the problem. Now the problem for you is, there are at most 9 trains in the station, all the trains has an ID(numbered from 1 to n), the trains get into the railway in an order O1, your task is to determine whether the trains can get out in an order O2.   

Input

The input contains several test cases. Each test case consists of an integer, the number of trains, and two strings, the order of the trains come in:O1, and the order of the trains leave:O2. The input is terminated by the end of file. More details in the Sample Input. 

Output

The output contains a string "No." if you can't exchange O2 to O1, or you should output a line contains "Yes.", and then output your way in exchanging the order(you should output "in" for a train getting into the railway, and "out" for a train getting out of the railway). Print a line contains "FINISH" after each test case. More details in the Sample Output. 

Sample Input

3 123 321

3 123 312

Sample Output

Yes.
in
in
in
out
out
out
FINISH
No.
FINISH


        
  
For the first Sample Input, we let train 1 get in, then train 2 and train 3.
So now train 3 is at the top of the railway, so train 3 can leave first, then train 2 and train 1.
In the second Sample input, we should let train 3 leave first, so we have to let train 1 get in, then train 2 and train 3.
Now we can let train 3 leave.
But after that we can't let train 1 leave before train 2, because train 2 is at the top of the railway at the moment.
So we output "No.".

这是很经典的问题了。学数据结构的时候，经常考。就是根据一个压栈的先后顺序，确定出栈的顺序。压栈序列我选择一个一个的判断，每次都先入栈，然后不断检验当前栈顶和出栈序列的首字母是否相同，在这里，出栈序列就是一个队列。所以直接用 stack 和 queue 就可以模拟。

易错就在检验栈顶和队列首相同后，应该还要继续检验，例子：入栈序列 1234 出栈序列 3241，发现 3 相同后，其实 2 也是相同的。

我通过储存 0 和 1 代表 In 和 out 。

#include <bits/stdc++.h>
using namespace std ;

int main(){
	// freopen( "Hdu 1022.txt" , "r" , stdin ) ;
	int n ;
	char In[12] , Out[12] ;
	while( ~scanf( "%d%s%s" , &n , In , Out ) ){
		string ans = "" ;
		stack<char> S ;
        queue<char> Q ;
		for( int i = 0 ; i < n ; ++i )
			Q.push( Out[i] ) ;
		for( int i = 0 ; i < n ; ++i ){
			S.push( In[i] ) ;
			ans += "0" ;
			while( !S.empty() && !Q.empty() && S.top() == Q.front() )
				ans += "1" , Q.pop() , S.pop() ;
		}
		while( !Q.empty() && !S.empty() ){
			if( Q.front() == S.top() )
				Q.pop() , S.pop() , ans += '1' ;
			else
				break ;
		}
		if( !S.empty() || !Q.empty() )
			printf( "No.\n" ) ;
		else{
			printf( "Yes.\n" ) ;
			for( int i = 0 ; i < (int)ans.size() ; ++i )
				printf( ans[i] == '1' ? "out\n" : "in\n" ) ;
		}
		printf( "FINISH\n" ) ;
	}
	return 0 ;
}