思路：
吧数字转化为子符串然后处理他的3个状态，用个map记录一下有没有访问过这个状态，并在node里记录路径
#pragma GCC optimize(2)
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include<iostream>
#include<vector>
#include<queue>
#include<map>
#include<stack>
#include<iomanip>
#include<cstring>
#include<time.h>
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define SIS std::ios::sync_with_stdio(false)
#define space putchar(' ')
#define enter putchar('\n')
#define lson root<<1
#define rson root<<1|1
typedef pair<int,int> PII;
const int mod=1e9+7;
const int N=2e6+10;
const int M=1e5+10;
const int inf=0x3f3f3f3f;
const int maxx=2e5+7;
const double eps=1e-6;
int gcd(int a,int b)
{
return b==0?a:gcd(b,a%b);
}
ll lcm(ll a,ll b)
{
return a*(b/gcd(a,b));
}
template <class T>
void read(T &x)
{
char c;
bool op = 0;
while(c = getchar(), c < '0' || c > '9')
if(c == '-')
op = 1;
x = c - '0';
while(c = getchar(), c >= '0' && c <= '9')
x = x * 10 + c - '0';
if(op)
x = -x;
}
template <class T>
void write(T x)
{
if(x < 0)
x = -x, putchar('-');
if(x >= 10)
write(x / 10);
putchar('0' + x % 10);
}
ll qsm(int a,int b,int p)
{
ll res=1%p;
while(b)
{
if(b&1)
res=res*a%p;
a=1ll*a*a%p;
b>>=1;
}
return res;
}
int n, m,k;
int dir[4][2]={
{
-1,0},{
0,1},{
1,0},{
0,-1}};
struct node
{
string chuan;
int step;
string apl;
};
string ed;
string st="12345678";
//int vis[1005][1005];
int sx,sy,ex,ey;
unordered_map<string ,int> mp;
queue<node> q;
string get1(string x)
{
string y=x;
for(int i=0;i<4;i++)
x[i]=y[8-i-1];
for(int i=4;i<8;i++)
x[i]=y[8-i-1];
return x;
}
string get2(string x)
{
string y=x;
x[0]=y[3];for(int i=1;i<4;i++) x[i]=y[i-1];
x[7]=y[4];for(int i=4;i<7;i++) x[i]=y[i+1];
return x;
}
string get3(string x)
{
string y=x;
x[1]=y[6];x[2]=y[1];x[5]=y[2];x[6]=y[5];
return x;
}
void bfs()
{
mp[st]=1;
q.push({
st,0,""});
while(!q.empty())
{
auto now=q.front();q.pop();
if(now.chuan==ed){
cout<<now.step<<endl;
cout<<now.apl<<endl;break;
}
string s1=get1(now.chuan);
if(!mp[s1]){
q.push({
s1,now.step+1,now.apl+"A"});
mp[s1]=1;
}
string s2=get2(now.chuan);
if(!mp[s2]){
q.push({
s2,now.step+1,now.apl+"B"});
mp[s2]=1;
}
string s3=get3(now.chuan);
if(!mp[s3]){
q.push({
s3,now.step+1,now.apl+"C"});
mp[s3]=1;
}
}
}
int main()
{
for(int i=0;i<8;i++)
{
int x;
scanf("%d",&x);
ed+=(char)(x+'0');
}
if(st==ed) cout<<0<<endl;
else
bfs();
return 0;
}