2013
12-04

# Open the Lock

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

题意：有一个四位数字的字符串，要求将初始数字串改变成目标数字串，允许的操作有两种：

1、将每一位加1或减1操作，1减1操作变成9，9加1操作变成1；

2、交换相邻的两个字符；

struct  Q

{

int x [4];// 每一位的数字

int steps;//当前状态已经花费的时间

}

bool vis[10000];//存储已经访问过的每一种状态，如果已经访问过此种状态就不在访问了。

#include<iostream>
#include<queue>
#include<cstdio>
using namespace std;
int s1,s2,s3,s4;
int e1,e2,e3,e4;
int steps;
bool vis[10000];
struct Q
{
int x[4];
int steps;
};
void bfs()
{
queue<Q>q;
Q temp,now;
int i,j;
int x[4];
int bit[4]={1000,100,10,1};

while(!q.empty())q.pop();
temp.x[0]=s1; temp.x[1]=s2;
temp.x[2]=s3; temp.x[3]=s4;
temp.steps=0;
q.push(temp);
steps=0xfffffff;
memset(vis,0,sizeof(vis));
vis[s1*1000+s2*100+s3*10+s4]=1;
while(!q.empty())
{
now=q.front();
q.pop();
if(now.x[0]==e1&&now.x[1]==e2&&now.x[2]==e3&&now.x[3]==e4)
{

if(steps>now.steps)
{
steps=now.steps;
}
}
for(i=0;i<4;i++)
{
for(j=0;j<4;j++)
{
x[0]=now.x[0];x[1]=now.x[1];
x[2]=now.x[2];x[3]=now.x[3];
temp=now;
if(j==0)
{
x[i]++;
if(x[i]>9)x[i]=1;
}
else
if(j==1)
{
x[i]--;
if(x[i]<1)x[i]=9;
}
else
if(j==2)
{
if(i<=0)continue;
int t=x[i];
x[i]=x[i-1];
x[i-1]=t;
}
else
{
if(i>=3)continue;
int t=x[i];
x[i]=x[i+1];
x[i+1]=t;
}
//printf("%d %d %d %d\n",x[0],x[1],x[2],x[3]);
if(vis[x[0]*bit[0]+x[1]*bit[1]+x[2]*bit[2]+x[3]*bit[3]])continue;
vis[x[0]*bit[0]+x[1]*bit[1]+x[2]*bit[2]+x[3]*bit[3]]=1;
temp.x[0]=x[0];temp.x[1]=x[1];temp.x[2]=x[2];temp.x[3]=x[3];

temp.steps++;
q.push(temp);
}
}
}
return;
}

int main()
{
int t;
char str[5];

scanf("%d",&t);
while(t--)
{
scanf("%s",str);
s1=str[0]-'0'; s2=str[1]-'0';
s3=str[2]-'0'; s4=str[3]-'0';
scanf("%s",str);
e1=str[0]-'0'; e2=str[1]-'0';
e3=str[2]-'0'; e4=str[3]-'0';
bfs();
printf("%d\n",steps);
}
return 0;
}