不带任何优化且使用了巨慢STL容器set来查重的随便在哪个OJ上提交都会TLE的八数码（基本是从刘汝佳抄的）

虽说这份代码的无能已经从题目看出来了，但是，他能打印步骤，他能打印步骤，他能打印步骤，重要的事情说三遍！（然并卵）

#include <iostream>
#include <set>
#include <cstring>
#include <vector>
#pragma GCC optimize(2)
#define _for(i,a,b) for(int i = (a);i < (b);i ++)
using namespace std;
typedef int State[9];
const int maxstate = 1000000;
State st[maxstate],goal;
int dist[maxstate] {0};
int fa[maxstate];

const int dx[] = {1,-1,0,0};
const int dy[] = {0,0,1,-1};

set<int> vis;
void init_lookup_table()
{
    vis.clear();
}
int try_to_insert(int s)
{
    int v = 0;
    _for(i,0,9) v = v*10+st[s][i];
    if(vis.count(v)) return 0;
    vis.insert(v);
    return 1;
}

bool limit(int newx,int newy)
{
    return newx>=0&&newx<3 && newy>=0&&newy<3;
}

int bfs()
{
    init_lookup_table();
    int front = 1,rear = 2;
    while(front<rear)
    {
        State &s = st[front];
        if(memcmp(goal,s,sizeof(s))==0)    return front;
        int z;
        for(z = 0; z < 9; z ++)
            if(!s[z]) break;//find zero
        int x = z/3,y = z%3;
        _for(d,0,4)
        {
            int newx = x+dx[d];
            int newy = y+dy[d];
            int newz = newx*3+newy;
            if(limit(newx,newy))
            {
                State &t = st[rear];//new node
                memcpy(&t,&s,sizeof(s));
                t[newz] = s[z];//0 move
                t[z] = s[newz];//move to where ori is 0
                dist[rear] = dist[front]+1;
                if(try_to_insert(rear))
                {
                    fa[rear] = front;
                    rear ++;
                }
            }
        }
        front ++;
    }
    return 0;
}

void print_ans(int u)
{
    vector<vector<int>> nodes;
    vector<int> tmp;
    _for(j,0,9)
    {
        tmp.push_back(st[u][j]);
    }
    nodes.push_back(tmp);
    while(1)
    {
        tmp.clear();
        if(fa[u]==0) break;
        _for(j,0,9)
        {
            tmp.push_back(st[fa[u]][j]);
        }
        nodes.push_back(tmp);
        u = fa[u];
    }
    int cnt = 0;
    int kase = 0;
    for(int i = nodes.size()-1; i >= 0; i --)
    {
        if(!kase)
        {
            cout << "Ori:" << endl;
            kase ++;
        }
        else
            cout << "Step:" << kase++ << endl;
        cout << nodes[i][0] << " " << nodes[i][1] << " " << nodes[i][2] << endl;
        cout << nodes[i][3] << " " << nodes[i][4] << " " << nodes[i][5] << endl;
        cout << nodes[i][6] << " " << nodes[i][7] << " " << nodes[i][8] << endl;
        cout << endl << endl;
    }
}

int main()
{
    //freopen("output.txt", "w", stdout);
    _for(i,0,9)    scanf("%d",&st[1][i]);
    _for(i,0,9) scanf("%d",&goal[i]);
    int ans = bfs();
    if(ans>0)
    {
        printf("Step Needed:%d times
",dist[ans]);
        print_ans(ans);
    }
    else printf("-1
");
    return 0;
}