HDU--杭电--1195--Open the Lock--深搜--忘记说句话易做图了，都是什么双向广搜，不知道怎么想的，直接就是一个深搜的水题好不好？

这个题我看了，都是推荐的神马双向广搜，难道这个深搜你们都木有发现？还是特意留个机会给我易做图？
Open the Lock
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3014 Accepted Submission(s): 1323

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

#include <iostream>
#include <cstring>
#include <cmath>
using namespace std;
int visit[4]={0},ss,a[4],b[4]; //visit记录a[i]是否被访问，ss用来记录暴力过程中的最小值
int cmp(int s) //这个是用来记录从当前这样的状态不左右移动最少要多少次
{
    int i,sum,aa[4]={s/1000,s/100%10,s/10%10,s%10}; //因为进来的是一个数字，所以用数组把它阉了，嘿嘿
    for(i=sum=0;i<4;i++) //一位一位的开锁
    {
        if(abs(aa[i]-b[i])<5)sum+=abs(aa[i]-b[i]);
        else sum+=9-abs(aa[i]-b[i]);
    }return sum;
}
void dfs(int sum,int s) //dfs里面进来的sum是当前的这个合成的数字，也就是不停移动得到的数组，s是记录移动次数
{
    int i;
    if(sum>999) //当sum>999也就是说sum是一个四位数时说明新的那个开锁初始状态OK了
    {
        i=cmp(sum)+s; //这时把sum和数组b开锁的次数和用s记录的移动的次数记录下来
        if(i<ss)ss=i; //比较，ss记录最小开锁操作
        return;
    }
    for(i=0;i<4;i++)
    {
        if(!visit[i])
        {
            visit[i]=1;
            dfs(sum*10+a[i],s++); //要是你聪明，会看到这里的一个重要的步骤，就是s++，别小看了，这说明下一次再在for循环里面取数就是在这次的数移动一位得到的，本体就是这个才是重点呢
            visit[i]=0;
        }
    }
}
int main (void)
{
    int i,j,k,l,n,m,t;
    cin>>t;
    while(t--&&cin>>n>>m)
    {
        for(i=3,j=1;i>=0;i--,j*=10)
        {
            a[i]=n/j%10;
            b[i]=m/j%10;
        }
        ss=99999;
        dfs(0,0);
        cout<<ss<<endl;

}
return 0;
}

补充：软件开发 , C++ ,