hdu-1195-Open the Lock

題目大意：一個四位數，可加一，9+1=1，可減一，1-1=9，可左右互換，但最左和最右不能換，要多少步換到第二個四位數。

廣搜拓展一下，向三個方向。

Open the Lock
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3055    Accepted Submission(s): 1346

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 <cstdio>   
#include <queue>   
#include <cstring>   
using namespace std;  
struct nobe  
{  
    int num[4],step;  
};  
  
queue<nobe> q;  
int visit[15][15][15][15];  
int x,y;  
void BFS(int x)  
{  
    nobe que,ue;  
    int i;  
    for(i=0; i<4; i++)  
    {  
        que.num[3-i]=x%10;  
        x/=10;  
    }  
    que.step=0;  
    visit[que.num[0]][que.num[1]][que.num[2]][que.num[3]]=1;  
    q.push(que);  
    while(!q.empty())  
    {  
        que=q.front();  
        q.pop();  
        if(que.num[0]*1000+que.num[1]*100+que.num[2]*10+que.num[3]==y)  
        {  
           printf("%d\n",que.step);  
        }  
        for(i=0; i<4; i++)//加一   
        {  
            ue=que;  
            if(que.num[i]==9)  
                ue.num[i]=1;  
            else  
                ue.num[i]=ue.num[i]+1;  
            if(!visit[ue.num[0]][ue.num[1]][ue.num[2]][ue.num[3]])  
            {  
                visit[ue.num[0]][ue.num[1]][ue.num[2]][ue.num[3]]=1;  
                ue.step=que.step+1;  
                q.push(ue);  
            }  
        }  
        for(i=0; i<4; i++)//減一   
        {  
            ue=que;  
            if(que.num[i]==1)  
                ue.num[i]=9;  
            else  
                ue.num[i]=ue.num[i]-1;  
            if(!visit[ue.num[0]][ue.num[1]][ue.num[2]][ue.num[3]])  
            {  
                visit[ue.num[0]][ue.num[1]][ue.num[2]][ue.num[3]]=1;  
                ue.step=que.step+1;  
                q.push(ue);  
            }  
        }  
        for(i=0; i<3; i++) //互換   
        {  
            ue=que;  
            ue.num[i]=que.num[i+1];  
            ue.num[i+1]=que.num[i];  
            if(!visit[ue.num[0]][ue.num[1]][ue.num[2]][ue.num[3]])  
            {  
                visit[ue.num[0]][ue.num[1]][ue.num[2]][ue.num[3]]=1;  
                ue.step=que.step+1;  
                q.push(ue);  
            }  
        }  
  
    }  
}  
int main()  
{  
    int i,k;  
    cin>>k;  
    for(i=0; i<k; i++)  
    {  
        memset(visit,0,sizeof(visit));  
        if(i)  
        getchar();  
        cin>>x>>y;  
        BFS(x);  
    }  
  
}  

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
struct nobe
{
    int num[4],step;
};

queue<nobe> q;
int visit[15][15][15][15];
int x,y;
void BFS(int x)
{
    nobe que,ue;
    int i;
    for(i=0; i<4; i++)
    {
        que.num[3-i]=x%10;
        x/=10;
    }
    que.step=0;
    visit[que.num[0]][que.num[1]][que.num[2]][que.num[3]]=1;
    q.push(que);
    while(!q.empty())
    {
        que=q.front();
        q.pop();
        if(que.num[0]*1000+que.num[1]*100+que.num[2]*10+que.num[3]==y)
        {
           printf("%d\n",que.step);
        }
        for(i=0; i<4; i++)//加一
        {
            ue=que;
            if(que.num[i]==9)
                ue.num[i]=1;
            else
                ue.num[i]=ue.num[i]+1;
            if(!visit[ue.num[0]][ue.num[1]][ue.num[2]][ue.num[3]])
            {
                visit[ue.num[0]][ue.num[1]][ue.num[2]][ue.num[3]]=1;
                ue.step=que.step+1;
                q.push(ue);
            }
        }
        for(i=0; i<4; i++)//減一
        {
            ue=que;
            if(que.num[i]==1)
                ue.num[i]=9;
            else
                ue.num[i]=ue.num[i]-1;
            if(!visit[ue.num[0]][ue.num[1]][ue.num[2]][ue.num[3]])
            {
                visit[ue.num[0]][ue.num[1]][ue.num[2]][ue.num[3]]=1;
                ue.step=que.step+1;
                q.push(ue);
            }
        }
        for(i=0; i<3; i++) //互換
        {
            ue=que;
            ue.num[i]=que.num[i+1];
            ue.num[i+1]=que.num[i];
            if(!visit[ue.num[0]][ue.num[1]][ue.num[2]][ue.num[3]])
            {
                visit[ue.num[0]][ue.num[1]][ue.num[2]][ue.num[3]]=1;
                ue.step=que.step+1;
                q.push(ue);
            }
        }

    }
}
int main()
{
    int i,k;
    cin>>k;
    for(i=0; i<k; i++)
    {
        memset(visit,0,sizeof(visit));
        if(i)
        getchar();
        cin>>x>>y;
        BFS(x);
    }

}