最小步数模型

#include <cstdio>
#include <algorithm>
#include <string>
#include <unordered_map>
#include <iostream>
#include <queue>
#include <map>
using namespace std;
queue<string> q;
char g[2][4];
unordered_map<string, int> dist;
unordered_map<string, pair<char, string>> pre;
void set(string state) //把传入的字符串变成2 * 4的矩阵
{
    for (int i = 0; i < 4; i ++ ) g[0][i] = state[i];
    for (int i = 7, j = 0; j < 4; j ++, i -- ) g[1][j] = state[i];
}
string get()            //把矩阵变成字符串
{
    string res;
    for (int i = 0; i < 4; i ++ ) res += g[0][i];
    for (int i = 3; i >= 0; i -- ) res += g[1][i];
    return res;
}
string move0(string state)
{
    set(state);
    for (int i = 0; i < 4; i ++ ) swap(g[0][i], g[1][i]);
    return get();
}
string move1(string state)
{
    set(state);
    char v0 = g[0][3], v1 = g[1][3];
    for (int i = 3; i > 0; i -- )
    {
        g[0][i] = g[0][i - 1];
        g[1][i] = g[1][i - 1];
    }
    g[0][0] = v0;
    g[1][0] = v1;
    return get();
}
string move2(string state)
{
    set(state);
    char v = g[0][1];
    g[0][1] = g[1][1];
    g[1][1] = g[1][2];
    g[1][2] = g[0][2];
    g[0][2] = v;
    return get();
}
int bfs(string start, string end)
{
    if (start == end) return 0;
    q.push(start);
    dist[start] = 0;
    while (!q.empty())
    {
        auto t = q.front();
        q.pop();
        string m[3];
        m[0] = move0(t);
        m[1] = move1(t);
        m[2] = move2(t);
        for (int i = 0; i < 3; i ++ )
            if (!dist.count(m[i]))
            {
                dist[m[i]] = dist[t] + 1;
                pre[m[i]] = {'A' + i, t};
                q.push(m[i]);
                if (m[i] == end) return dist[end];
            }
    }
    return -1;
}
int main()
{
    int x;
    string start, end;
    for (int i = 0; i < 8; i ++ ) 
    {
        scanf("%d", &x);
        end += x + '0';
    }
    for (int i = 1; i <= 8; i ++ ) start += i + '0';
    int step = bfs(start, end);
    printf("%d\n", step);
    string res;
    while (start != end)
    {
        res += pre[end].first;
        end = pre[end].second;
    }
    reverse(res.begin(), res.end());
    if (step > 0) cout << res;
    return 0;
}