hdu 1195 Open the Lock

Open the Lock

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4534    Accepted Submission(s): 1993

Problem Description

Now an emergent task for you is to open a password lock. The password is consisted of four digits. Each digit is numbered from 1 to 9. 
Each time, you can add or minus 1 to any digit. When add 1 to '9', the digit will change to be '1' and when minus 1 to '1', the digit will change to be '9'. You can also exchange the digit with its neighbor. Each action will take one step.

Now your task is to use minimal steps to open the lock.

Note: The leftmost digit is not the neighbor of the rightmost digit.

 

 

Input

The input file begins with an integer T, indicating the number of test cases. 

Each test case begins with a four digit N, indicating the initial state of the password lock. Then followed a line with anotther four dight M, indicating the password which can open the lock. There is one blank line after each test case.

 

 

Output

For each test case, print the minimal steps in one line.

 

 

Sample Input

2

1234

2144

1111

9999

 

Sample Output

2

4

 

Author

YE, Kai

 

Source

Zhejiang University Local Contest 2005

 

 

PS：思路来自其他大神

还是功力不足啊，针对一个数，就只有三种情况，+1（4），-1（4），与邻位交换（3），也就是说一个4位数，动一次可以有12种变化

 

  1 #include<iostream>  2 #include<cstdio>  3 #include<cmath>  4 #include<cstdlib>  5 #include<cstring>  6 #include<queue>  7 using namespace std;  8 struct node{  9     int num[6]; 10     int step; 11 }; 12 node first,last; 13 bool vis[10][10][10][10],flag; 14  15 int bfs() 16 { 17     node p,q; 18     int i; 19     memset(vis,false,sizeof(vis)); 20     queue<struct node>Q; 21     first.step=0; 22     p=first; 23     vis[p.num[0]][p.num[1]][p.num[2]][p.num[3]]=true; 24     Q.push(p); 25     while(!Q.empty()) 26     { 27         flag=true; 28         p=Q.front(); 29         Q.pop(); 30         for(i=0;i<4;i++) 31         { 32             if(p.num[i]!=last.num[i]) 33             { 34                 flag=false; 35                 break; 36             } 37         } 38         if(flag) 39         { 40             return p.step; 41         } 42         for(i=0;i<4;i++)//加法 43         { 44             q=p; 45             if(p.num[i]==9) 46             q.num[i]=1; 47             else 48             q.num[i]=p.num[i]+1; 49             q.step=p.step+1; 50             if(!vis[q.num[0]][q.num[1]][q.num[2]][q.num[3]]) 51             { 52                 vis[q.num[0]][q.num[1]][q.num[2]][q.num[3]]=true; 53                 Q.push(q); 54             } 55         } 56         for(i=0;i<4;i++)//减法 57         { 58             q=p; 59             if(p.num[i]==1) 60             q.num[i]=9; 61             else 62             q.num[i]=p.num[i]-1; 63             q.step=p.step+1; 64             if(!vis[q.num[0]][q.num[1]][q.num[2]][q.num[3]]) 65             { 66                 vis[q.num[0]][q.num[1]][q.num[2]][q.num[3]]=true; 67                 Q.push(q); 68             } 69         } 70         for(i=0;i<3;i++)//邻位交换 71         { 72             q=p; 73             if(q.num[i+1]!=p.num[i]) 74             { 75                 q.num[i+1]=p.num[i]; 76                 q.num[i]=p.num[i+1]; 77             } 78             q.step=p.step+1; 79             if(!vis[q.num[0]][q.num[1]][q.num[2]][q.num[3]]) 80             { 81                 vis[q.num[0]][q.num[1]][q.num[2]][q.num[3]]=true; 82                 Q.push(q); 83             } 84         } 85     } 86     return 0; 87 } 88  89  90 int main() 91 { 92     int n; 93     char s1[6],s2[6]; 94     scanf("%d",&n); 95     while(n--) 96     { 97         int i; 98         scanf("%s%s",s1,s2); 99         for(i=0;i<4;i++)100         {101             first.num[i]=s1[i]-'0';102             last.num[i]=s2[i]-'0';103         }104         int count;105         count=bfs();106         printf("%d\n",count);107 108     }109     return 0;110 }

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