Open the Lock

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 3825    Accepted Submission(s): 1650

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

题目：http://acm.hdu.edu.cn/showproblem.php?pid=1195

题意就是开锁，四个数字的锁，1~9没有0哟。

每次可以对任意一个数字+1或者-1，当数字为1，减1，则数字变为9。

同样，当数字为9，若+1，数字则变为1。

每次操作除了对任意一位数字+1或者-1之外，还可以对相邻位置的数字进行交换。

第一位置与第二位置 or 第二位置和第三位置 or 第三位置和第四位置

但是第一位置和第四位置不可以交换。

暴力广搜，尝试过剪枝，但发现减了就WA。。。o(╯□╰)o。。。

暴力也才62MS，等一会做个双广的来试试。

双向BFS已A，详情请戳：http://blog.csdn.net/lttree/article/details/24669525

/**************************************
***************************************
*        Author：Tree                 *
*From  ：http://blog.csdn.net/lttree  *
* Title : Scrambled Polygon           *
*Source: hdu 1195                     *
* Hint  : 暴力BFS                     *
***************************************
**************************************/

#include <iostream>
#include <string.h>
#include <queue>
using namespace std;
struct Key
{
    int k[4],step;
};
bool vis[10001];         // 9^4
int ini[4],ans[4],dis[2]={-1,1};
int solu[9]={9,1,2,3,4,5,6,7,8};

bool judge(Key a)
{
    for(int i=0;i<4;++i)
        if( a.k[i]!=ans[i] )
            return false;
    return true;
}
// 用来做VIS数组
int total(Key a)
{
    int i,sum;
    sum=0;
    for(i=0;i<4;++i)
        sum=sum*10+a.k[i];
    return sum;
}
int bfs(void)
{
    memset(vis,0,sizeof(vis));
    queue <Key> q;
    Key pre,lst;
    int i,t,sum;

    for(i=0;i<4;++i)
        pre.k[i]=ini[i];
    sum=total(pre);
    vis[sum]=1;
    pre.step=0;
    q.push(pre);

    while( !q.empty() )
    {
        pre=q.front();
        q.pop();
        if( judge(pre) )    return pre.step;

        // 每位数 加1或减1（若相应位置与答案不同）
        for(i=0;i<4;++i)
        {
            if( pre.k[i]!=ans[i] )
            {
                for(t=0; t<2; ++t)
                {
                    lst=pre;
                    lst.k[i]=solu[(pre.k[i]+dis[t])%9];
                    sum=total(lst);
                    if( vis[sum] )  continue;
                    lst.step=pre.step+1;
                    vis[sum]=1;
                    q.push(lst);
                }
            }
        }

        // 相邻交换
        for(i=0;i<3;++i)
        {
            lst=pre;
            lst.k[i]=pre.k[i+1];
            lst.k[i+1]=pre.k[i];
            sum=total(lst);
            if( !vis[sum] )
            {
                lst.step=pre.step+1;
                vis[sum]=1;
                q.push(lst);
            }
        }

    }
}
int main()
{
    int i,s,test;
    char c;
    cin>>test;
    while( test-- )
    {
        // 输入数据
        for(i=0;i<4;++i)
        {
            cin>>c;
            ini[i]=c-‘0‘;
        }
        for(i=0;i<4;++i)
        {
            cin>>c;
            ans[i]=c-‘0‘;
        }

        s=bfs();
        cout<<s<<endl;
    }
    return 0;
}