zoukankan      html  css  js  c++  java
  • Train Problem I(HDU1022)

    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
    //这是简单的栈应用,先进先出原则。

    #include <stdio.h>
    #include <string.h>
    int main
    ()
    {
     int n,i,j,k,t,r[23],stack[13];
      char
     si[13],so[13];
      while
    (scanf("%d",&n)!=EOF)
      {

          getchar();
          memset(r,0,sizeof(r));
          memset(stack,0,sizeof(stack));
          scanf("%s%s",si,so);
          stack[1]=si[0];r[0]=1;
          for
    (k=t=j=1,i=0;so[i]!='\0';i++)
           {

              while
    (t<n)
              {

                if
    (j>0&&stack[j]==so[i]) //出栈
                {

                   j--;
                  r[k++]=2;//记录进出栈的情况
                  break
    ;
                }

                stack[++j]=si[t++];//进栈
                r[k++]=1; //记录进出栈的情况
              }

             if
    (stack[j]==so[i])//出栈
            {

                j--;
                r[k++]=2; //记录进出栈的情况
            }
         }


           n=n<<1;
           if
    (k<n)
             printf("No.\nFINISH\n");//这里开始比较郁闷的,No写成了大写还有下面的Yes,唉悲剧!
           else

             {
       printf("Yes.\n");
                 for
    (i=0;i<k;i++)
                   switch
    (r[i])
                   {

                       case
     1:printf("in\n");break;
                       case
     2:printf("out\n");break;
                   }

                 printf("FINISH\n");
             }
      }

      return
     0;
    }

                                                          -----------江财小子

  • 相关阅读:
    OpenIOC
    网站舆情监测
    乌云的背后是阳光
    2014 十大工具
    NetFlow网络流量监测技术的应用和设计(转载)
    免费工具
    Oracle RAC环境下怎样更新patch(Rolling Patch)
    Answer&#39;s Question about pointer
    cocos2d-x 3.0 final 移植 android
    ReactNavtive框架教程(3)
  • 原文地址:https://www.cnblogs.com/372465774y/p/2421641.html
Copyright © 2011-2022 走看看