hdu 1022火车进出站问题

Posted 2020-10-14 xzy不会飞的地板流

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.

 

 

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

 

题意：

 

  1 import java.util.Scanner;
  2 import java.util.Stack;
  3 
  4 public class Test5 {
  5     
  6      static Stack<Character> s;
  7      static char[] in,out;
  8      static String[] res;
  9      static String str1,str2;
 10      static int n;
 11      static int count1,count2;
 12 
 13     public static void main(String[] args) {
 14 
 15         Scanner input=new Scanner(System.in);
 16         
 17         while(input.hasNext()) {
 18             
 19             n=input.nextInt();
 20             
 21             s=new Stack<Character>();
 22             in=new char[n+1];//火车进站顺序
 23             out=new char[n+1];//火车出站顺序
 24             res=new String[2*n+1];//每个火车进出站记录
 25             
 26             str1=input.next();//火车进站序列
 27             str2=input.next();//火车出站序列
 28             
 29             for(int i=0;i<n;i++) {
 30                 
 31                 in[i]=str1.charAt(i);
 32                 out[i]=str2.charAt(i);
 33                 
 34             }
 35             
 36             count1=0;//进站的列车序号
 37             count2=0;//出站的列车序号
 38             boolean istrue=dfs(0);
 39             
 40             if(istrue) {
 41                 
 42                 System.out.println("Yes.");
 43                 for(int i=0;i<2*n;i++) {
 44                     
 45                     System.out.println(res[i]);
 46                     
 47                 }
 48                 
 49             }else {
 50                 
 51                 System.out.println("No.");
 52                 
 53             }
 54             
 55             System.out.println("FINISH");
 56             
 57         }
 58 
 59     }
 60 
 61     //判断是否可以按str2顺序出站
 62     private static boolean dfs(int k) {
 63         
 64         if(count2==n) {//若出站列车序号为n时，说明前面所有的列车都可按照str2顺序出站
 65             
 66             return true;
 67             
 68         }
 69         
 70         if(k==n&&count2<n) {//当进站列车序号为n，说名所有火车都已经进站，count2<n说明出站火车列表中还有火车没有出站
 71             
 72             return false;
 73             
 74         }
 75         
 76         if(in[k]!=out[count2]) {//若序号为k的进站列车和出站列车序号不相等，则让序号为k的进站列车进站；然后判断下一个进站列车情况
 77             
 78             s.push(in[k]);
 79             res[count1++]="in";
 80             return dfs(k+1);
 81             
 82         }else {//(1)若序号为k的进站列车与序号为count2的出站列车序号相等,则让k先进站，然后立即出站；并将count2++(下一个该出站的列车)
 83             
 84             //(2)然后判断栈定列车序号是否和出站列车序号是否相等，若相等将栈顶列车出站并做记录，然后将count2++(下一个该出站的列车),重复步骤2直到栈为空或者栈顶火车与count2(下一个该出站的火车)不相等，否则判断下一个火车情况
 85             res[count1++]="in";
 86             res[count1++]="out";
 87             count2++;
 88             while(!s.isEmpty()) {
 89                 
 90                 if(s.peek()!=out[count2]) {
 91                     
 92                     break;
 93                     
 94                 }
 95                 
 96                 res[count1++]="out";
 97                 s.pop();
 98                 count2++;
 99             }
100             
101             return dfs(k+1);
102             
103         }
104         
105         
106     }
107 
108 }