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.

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.

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

2
1234
2144

1111
9999

2
4

YE, Kai

Zhejiang University Local Contest 2005

Ignatius.L

知道是BFS 的题目，但想不到怎么去搜，据说比素数环还要简单，经过别人的点拨和网上的代码，瞬间发现，想通了真的比素数环还要简单

#include <iostream>
#include <queue>
#include <cstring>
using namespace std;
struct node
{
    int a[4],step;
};
char str[2][5];
int f[4];
bool vis[10][10][10][10];
int BFS()
{
    node now,next;
    for(int i=0;i<4;i++)
    {
        now.a[i]=str[0][i]-'0';
        f[i]=str[1][i]-'0';
    }
    memset(vis,false,sizeof(vis));
    now.step=0;
    queue<node>q;
    q.push(now);
    bool lock;

    while(!q.empty())
    {
        now=q.front();
        q.pop();
        lock=true;
        for(int i=0;i<4;i++)
        {
            if(now.a[i]!=f[i])
            {
                lock=false;
                break;
            }
        }
        if(lock)
            return now.step;
        for(int i=0;i<4;i++)//加一
        {
            next=now;
            if(now.a[i]==9)
                next.a[i]=1;
            else
                next.a[i]=now.a[i]+1;
            if(!vis[next.a[0]][next.a[1]][next.a[2]][next.a[3]])
            {
                vis[next.a[0]][next.a[1]][next.a[2]][next.a[3]]=true;
                next.step=now.step+1;
                q.push(next);
            }
        }
        for(int i=0;i<4;i++)//减一
        {
            next=now;
            if(now.a[i]==1)
                next.a[i]=9;
            else
                next.a[i]=now.a[i]-1;
            if(!vis[next.a[0]][next.a[1]][next.a[2]][next.a[3]])
            {
                vis[next.a[0]][next.a[1]][next.a[2]][next.a[3]]=true;
                next.step=now.step+1;
                q.push(next);
            }
        }
        for(int i=0;i<3;i++)//交换
        {
            next=now;
            next.a[i]=now.a[i+1];
            next.a[i+1]=now.a[i];
            if(!vis[next.a[0]][next.a[1]][next.a[2]][next.a[3]])
            {
                vis[next.a[0]][next.a[1]][next.a[2]][next.a[3]]=true;
                next.step=now.step+1;
                q.push(next);
            }
        }
    }
    return -1;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        cin>>str[0]>>str[1];
        cout<<BFS()<<endl;
    }
    return 0;
}