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HDU 1022Train Problem I

Train Problem I

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 22970    Accepted Submission(s): 8659

Problem Description 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.

HDU 1022Train Problem I
HDU 1022Train Problem I
HDU 1022Train Problem I

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


   
    
     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.".
        

Author Ignatius.L

給你一些火車,以數字來進行标号,判斷以一的方式進站,是否能以二的方式出站。假如1 2 3 進站 則可以 3 2 1 出站。 假如1 3 2 進站 也可以 1 2 3出站,因為1進去後可以先出來。然後 2 3 進。 3 2 出。棧主要就是先進後出的思想。

上代碼、

#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 類型的放進隊列。
    int t;
    while(~scanf("%d%s%s",&t,s1,s2))
    {
        i=j=wbx=0;
        while(!s.empty())  
            s.pop();   //   清空棧。
        while(i<t)
        {
            s.push(s1[i++]);   //    開始進站。
                vis[wbx++]=1;  //    将進站的in存入數組,令為一。
            while(!s.empty() &&s.top()==s2[j])//   如果棧不空的話,并且棧首元素等于出站元素。則彈棧。
            {
                j++;
                vis[wbx++]=0;//  出站的out也存入數組,令為0。
                s.pop();//  彈棧。
            }
        }
        if(!s.empty())
            printf("No.\n");//   上面做好之後,如果此時棧還不是空的,則不能以方式2的順序出站。
        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;
}