HDU 1195 Open the Lock（BFS）

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. 

InputThe 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. 
OutputFor each test case, print the minimal steps in one line. 
Sample Input

2
1234
2144

1111
9999

Sample Output

2
4
题意：
　　　给出两组数字，问最短需要几步可以变换到想要的数字。
题解：
　　　刚开始是没有什么思路的，感觉很麻烦。其实想通了之后很简单，就是普通的BFS，只不过这道题的搜索方式很多，代码长了一些。

#include<iostream>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m,ans[4];
const int maxn=1e4+5;
struct node
{
    int a[4],step;
};
bool vis[10][10][10][10];
int bfs()
{
    queue<node> que;
    node cur;
    int temp=1000;
    for(int i=0;i<4;i++)
    {
        cur.a[i]=n/temp%10;
        ans[i]=m/temp%10;
        temp/=10;
    }
    cur.step=0;
    que.push(cur);
    while(que.size())
    {
        cur=que.front();
        que.pop();
        int k;
        for(k=0;k<4;k++)
            if(cur.a[k]!=ans[k])
                break;
        if(k==4)
            return cur.step;
        node next;
        for(int i=0;i<4;i++)//加一
        {
            next=cur;
            if(next.a[i]==9)
                next.a[i]=1;
            else
                next.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++;
                que.push(next);
            }
        }
        for(int i=0;i<4;i++)//减一
        {
            next=cur;
            if(next.a[i]==1)
                next.a[i]=9;
            else
                next.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++;
                que.push(next);
            }
        }
        for(int i=0;i<3;i++)//相邻数字交换
        {
            next=cur;
            swap(next.a[i],next.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++;
                que.push(next);
            }
        }
    }
    return -1;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        cin>>n>>m;
        memset(vis,false,sizeof(vis));
        cout<<bfs()<<endl;
    }
    return 0;
}