Train Problem I

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.

3 123 321
3 123 312

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

Hint
Hint
 
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.".

 栈的水题一道。就直接对进出栈进行模拟。和之前做的那道题类似，用来练练STL还是不错的，直接用STL水过。

http://blog.csdn.net/whjkm/article/details/38455291 能够看看之前的那道题。

用来vector 和 stack容器。用vector来存储字符串；

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <stack>
#include <vector>
using namespace std;
int main()
{
    int n,i,j;
    string str1,str2;
    while(cin>>n>>str1>>str2)
    {
        stack<int>s;
        vector<string>v;
        for(i=0,j=0;i<n&&j<n;i++)
        {
            s.push(str1[i]);//把元素进栈
            v.push_back("in");//压进vector容器
            while(s.top()==str2[j])//进行匹配
            {
                if(!s.empty())
                {
                    s.pop();
                    v.push_back("out");
                }
                j++;
                if(s.empty()) break;
            }
        }
        if(j==n)
        {
            printf("Yes.\n");
            for(i=0; i< v.size();i++)
             cout<< v[i] <<endl;
             printf("FINISH\n");
        }
        else
             printf("No.\nFINISH\n");
    }
    return 0;
}