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.
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.
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
思路：
BFS 可以有+1 -1 交换三种线路 进行BFS
BFS 可以有+1 -1 交换三种线路 进行BFS
参考代码：
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#include <iostream>#include <string>#include <queue>using
namespace
std;
char
s1[5],s2[5];
int
a[4],b[4];
int
bfs();
int
mark[10000];
typedef
struct
{ int
x[4];
int
step;
}node;int
main()
{ int
t;
cin>>t;
while(t--)
{
cin>>s1>>s2;
int
i;
for(i=0;i<4;i++)
{
a[i]=s1[i]-‘0‘;
b[i]=s2[i]-‘0‘;
}
cout<<bfs()<<endl;
}
return
0;
}int
bfs()
{ memset(mark,0,sizeof(mark));
node temp;
queue<node> q;
temp.step=0;
int
i;
for(i=0;i<4;i++)
temp.x[i]=a[i];
q.push(temp);
mark[a[0]*1000+a[1]*100+a[2]*10+a[3]]=1;
while(!q.empty())
{
for(i=0;i<11;i++)
{
if(i<4)//加
{
node newl=q.front();
newl.x[i]=q.front().x[i]+1;
if(newl.x[i]==10)
newl.x[i]=1;
if(!mark[newl.x[0]*1000+newl.x[1]*100+newl.x[2]*10+newl.x[3]])
{
mark[newl.x[0]*1000+newl.x[1]*100+newl.x[2]*10+newl.x[3]]=1;
newl.step=q.front().step+1;
q.push(newl);
if(newl.x[0]==b[0]&&newl.x[1]==b[1]&&newl.x[2]==b[2]&&newl.x[3]==b[3])
return
newl.step;
}
}
else
if(i<8)//减
{
node newl=q.front();
newl.x[i-4]=q.front().x[i-4]-1;
if(newl.x[i-4]==0)
newl.x[i-4]=9;
if(!mark[newl.x[0]*1000+newl.x[1]*100+newl.x[2]*10+newl.x[3]])
{
mark[newl.x[0]*1000+newl.x[1]*100+newl.x[2]*10+newl.x[3]]=1;
newl.step=q.front().step+1;
q.push(newl);
if(newl.x[0]==b[0]&&newl.x[1]==b[1]&&newl.x[2]==b[2]&&newl.x[3]==b[3])
return
newl.step;
}
}
else
{
node newl=q.front();
int
temp;
temp=q.front().x[i-7];
newl.x[i-8]=temp;
newl.x[i-7]=q.front().x[i-8];
if(!mark[newl.x[0]*1000+newl.x[1]*100+newl.x[2]*10+newl.x[3]])
{
mark[newl.x[0]*1000+newl.x[1]*100+newl.x[2]*10+newl.x[3]]=1;
newl.step=q.front().step+1;
q.push(newl);
if(newl.x[0]==b[0]&&newl.x[1]==b[1]&&newl.x[2]==b[2]&&newl.x[3]==b[3])
return
newl.step;
}
}
}
q.pop();
}
} |