Blue Bridge Cup 2017- provincial tournament -C / C ++ - A Group 2 title

Topic
Title: Jumping Grasshopper

As shown in FIG p1.png:
There are nine plates, arranged in a circle.
Grasshoppers 8 wherein the filled tray 8, there is a blank disc.
We call these grasshoppers clockwise numbered from 1 to 8

Each grasshopper can jump to an adjacent empty dish,
can then point force, grasshoppers jump across a neighboring empty tray.

Please you do the math, if you want to make grasshoppers their formation changed in accordance with the counter-clockwise order,
and to maintain the same position a blank disc (that is, transposition 1-8, 2-7 transposition, ...), at least how many times after the jump?

Note: Requires submission is an integer, please do not fill out any extra content or captions.

Thinking

bfs

Code

 1 #include<iostream>
 2 #include<string.h>
 3 #include<queue>
 4 #include<stdlib.h>
 5 #include<cmath>
 6 #include<stdio.h>
 7 #include<algorithm>
 8 #define maxx 1000000000
 9 using namespace std;
10 long long int str_t_int(char str[]){//str to int
11      long long int sum=0,ll=1;
12      for(int i=8;i>=0;i--){
13          sum=sum+((str[i]-'0')*ll); 
14          ll=ll*10;
15      }
16      return sum;
17 }
18 
19 int find_9(char str[]){
20      int loc_9;
21      for(int i=0;i<10;i++){
22          if(str[i]=='9') Loc_9 = I; // find subscript a blank disc 
23 is       }
 24       return loc_9;
 25  }
 26 is  
27  void int_t_str ( int NUM, char * STR) {
 28       int I = . 8 ;
 29       the while (NUM% 10 ) {
 30           * (STR + I) = ( char ) (NUM% 10 + ' 0 ' );
 31 is           NUM / = 10 ;
 32           i-- ;
 33 is       }
 34 is      return ;
 35  }
 36  int the dir [ . 4 ] = {- 2 , - . 1 , . 1 , 2 }; // hopping scheme 
37 [  BOOL index [Maxx]; // if it has been determined in each case marking 
38 is    
39  int BFS ( ) {
 40       Long  Long  int Start = 123456789 ;
 41 is       Long  Long  int End = 876 543 219 ;
 42 is       Queue < int > Q;
 43 is       Queue < int >QC;
 44 is       q.push (Start);
 45       qc.push ( . 1 );    
 46 is       char STR [ 10 ];
 47       Long  Long  int TEMP, loc_9, temp_char, CNT;
 48       int Find = 0 ;
 49       index [Start] = . 1 ;
 50       CNT = . 1 ;
 51 is       the while ( . 1 ) {
 52 is           TEMP = q.front (); // get the current q queue header 
53 is           CNT = qc.front (); // get the current qc queue header 
54 is            
55          int_t_str (TEMP, STR); // int to STR 
56 is           
57 is           loc_9 = find_9 (STR); // check position 9 
58           
59           for ( int I = 0 ; I < . 4 ; I ++) { // begin by one trial 
60           
61 is               STR = temp_char [loc_9]; // analog skip 
62 is               STR [loc_9] STR = [(+ loc_9 the dir [I] + . 9 )% . 9 ];
 63 is               STR [(+ loc_9 the dir [I] + . 9 )% . 9 ] = temp_char;
 64               
65               
66               TEMP = str_t_int (STR);// STR to int 
67              
68  
69          
70               IF (! Index [TEMP]) { // if repeated 
71 is               
72                   IF (TEMP == End) { // find the queue 
73 is                       return CNT;
 74                   } the else { // mark temp, while temp is pressed into the queue 
75                       index [temp] = . 1 ;
 76                       q.push (temp);
 77                       qc.push (CNT + . 1 );
 78                   }
 79               }
 80               
81              STR = temp_char [loc_9]; // return situ, exploratory other directions 
82               STR [loc_9] STR = [(+ loc_9 the dir [I] + . 9 )% . 9 ];
 83               STR [(+ loc_9 the dir [I] + . 9 )% . 9 ] = temp_char;    
 84           }
 85           q.pop (); // Q, pop queue QC 
86           qc.pop ();
 87           
88       }
 89       
90  }
 91 is  int main () {
 92       COUT << BFS ( ) << endl;
 93  
94 }