两个队列轮流处理正向反向，并且用vis和step记录是否被处理过以及到达此处所需要的步数。
因为数据量很小只有1e4因此不用hash记录也可以，当然用hash能优化一点常数时间。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <map>
#include <cmath>
#include <algorithm>
#define inf 0x3f3f3f3f
#define SI(a) scanf("%d",&a)
#define set0(a) memset(a,0,sizeof(a))
typedef long long ll;
const int mod = 998244353;
const int maxn = 10010;
using namespace std;
int ans;
string s,e;
int vis[maxn],step[maxn];//0:没有处理过，1：正向处理过，2：逆向处理过
void bfs()
{
string n1,n2,t1,t2,T1,T2;
int num1 = stoi(s),num2 = stoi(e);
//cout<<num1<<" "<<num2<<endl;
n1 = s;n2 = e;
vis[num1] = 1,vis[num2] = 2;
queue<string>q1,q2;
q1.push(n1),q2.push(n2);
while(!q1.empty()||!q2.empty())
{
if(!q1.empty())
{
t1 = q1.front();
//cout<<t1<<endl;
q1.pop();
int temp = step[stoi(t1)];
for(int i = 0;i < 4;i++)
{
T1 = t1;
if(T1[i] == '1') T1[i] = '9';
else T1[i] = T1[i] - 1;
//cout<<T1<<endl;
num1 = stoi(T1);
if(vis[num1] == 1)continue;
else if(vis[num1] == 2){
ans = min(ans,step[num1] + temp + 1);}
else{
vis[num1] = 1;
step[num1] = temp + 1;
q1.push(T1);
}
}
for(int i = 0;i < 4;i++)
{
T1 = t1;
if(T1[i] == '1') T1[i] = '9';
else T1[i] = T1[i] - 1;
//cout<<T1<<endl;
num1 = stoi(T1);
if(vis[num1] == 1)continue;
else if(vis[num1] == 2){
ans = min(ans,step[num1] + temp + 1);}
else{
vis[num1] = 1;
step[num1] = temp + 1;
q1.push(T1);
}
}
for(int i = 0;i < 3;i++)
{
T1 = t1;
char t = T1[i];
T1[i] = T1[i+1];
T1[i+1] = t;
num1 = stoi(T1);
if(vis[num1] == 1)continue;
else if(vis[num1] == 2){
ans = min(ans,step[num1] + temp + 1);}
else{
vis[num1] = 1;
step[num1] = temp + 1;
q1.push(T1);
}
}
}
if(!q2.empty())
{
t2 = q2.front();
q2.pop();
int temp = step[stoi(t2)];
for(int i = 0;i < 4;i++)
{
T2 = t2;
if(T2[i] == '1') T2[i] = '9';
else T2[i] = T2[i] - 1;
num2 = stoi(T2);
if(vis[num2] == 2)continue;
else if(vis[num2] == 1){
ans = min(ans,step[num2] + temp + 1);}
else{
vis[num2] = 2;
step[num2] = temp + 1;
q2.push(T2);
}
}
for(int i = 0;i < 4;i++)
{
T2 = t2;
if(T2[i] == '1') T2[i] = '9';
else T2[i] = T2[i] - 1;
num2 = stoi(T2);
if(vis[num2] == 2)continue;
else if(vis[num2] == 1){
ans = min(ans,step[num2] + temp + 1);}
else{
vis[num2] = 2;
step[num2] = temp + 1;
q2.push(T2);
}
}
for(int i = 0;i < 3;i++)
{
T2 = t2;
char t = T2[i];
T2[i] = T2[i+1];
T2[i+1] = t;
num2 = stoi(T2);
if(vis[num2] == 2)continue;
else if(vis[num2] == 1){
ans = min(ans,step[num2] + temp + 1);}
else{
vis[num2] = 2;
step[num2] = temp + 1;
q2.push(T2);
}
}
}
}
}
int main()
{
int r;
SI(r);
while(r--)
{
set0(vis);
memset(step,0,sizeof(step));
ans = inf;
cin>>s>>e;
if(s == e)cout<<0<<endl;
else{
bfs();
cout<<ans<<endl;
}
}
return 0;
}