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.".

给你一些火车，以数字来进行标号，判断以一的方式进站，是否能以二的方式出站。假如1 2 3 进站 则可以 3 2 1 出站。 假如1 3 2 进站 也可以 1 2 3出站，因为1进去后可以先出来。然后 2 3 进。 3 2 出。栈主要就是先进后出的思想。

上代码、

#include <stdio.h>
#include <stack>
#include <string.h>
#include <algorithm>
using namespace std;
int main()
{
    int i,j,wbx;
    int vis[500];
    char s1[1000];
    char s2[1000];
    stack<char>s;  //将char 类型的放进队列。
    int t;
    while(~scanf("%d%s%s",&t,s1,s2))
    {
        i=j=wbx=0;
        while(!s.empty())  
            s.pop();   //   清空栈。
        while(i<t)
        {
            s.push(s1[i++]);   //    开始进站。
                vis[wbx++]=1;  //    将进站的in存入数组，令为一。
            while(!s.empty() &&s.top()==s2[j])//   如果栈不空的话，并且栈首元素等于出站元素。则弹栈。
            {
                j++;
                vis[wbx++]=0;//  出站的out也存入数组，令为0。
                s.pop();//  弹栈。
            }
        }
        if(!s.empty())
            printf("No.\n");//   上面做好之后，如果此时栈还不是空的，则不能以方式2的顺序出站。
        else //反之可以。
        {
            printf("Yes.\n");
            for(i=0;i<wbx;i++)
            {
                if(vis[i])
                    printf("in\n");
                else
                    printf("out\n");
            }
        }
        printf("FINISH\n");//  完成
    }
    return 0;
}

HDU 1022