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）
1 #include <iostream> 2 #include <set> 3 #include <cstring> 4 #include <vector> 5 #pragma GCC optimize(2) 6 #define _for(i,a,b) for(int i = (a);i < (b);i ++) 7 using namespace std; 8 typedef int State[9]; 9 const int maxstate = 1000000; 10 State st[maxstate],goal; 11 int dist[maxstate] {0}; 12 int fa[maxstate]; 13 14 const int dx[] = {1,-1,0,0}; 15 const int dy[] = {0,0,1,-1}; 16 17 set<int> vis; 18 void init_lookup_table() 19 { 20 vis.clear(); 21 } 22 int try_to_insert(int s) 23 { 24 int v = 0; 25 _for(i,0,9) v = v*10+st[s][i]; 26 if(vis.count(v)) return 0; 27 vis.insert(v); 28 return 1; 29 } 30 31 bool limit(int newx,int newy) 32 { 33 return newx>=0&&newx<3 && newy>=0&&newy<3; 34 } 35 36 int bfs() 37 { 38 init_lookup_table(); 39 int front = 1,rear = 2; 40 while(front<rear) 41 { 42 State &s = st[front]; 43 if(memcmp(goal,s,sizeof(s))==0) return front; 44 int z; 45 for(z = 0; z < 9; z ++) 46 if(!s[z]) break;//find zero 47 int x = z/3,y = z%3; 48 _for(d,0,4) 49 { 50 int newx = x+dx[d]; 51 int newy = y+dy[d]; 52 int newz = newx*3+newy; 53 if(limit(newx,newy)) 54 { 55 State &t = st[rear];//new node 56 memcpy(&t,&s,sizeof(s)); 57 t[newz] = s[z];//0 move 58 t[z] = s[newz];//move to where ori is 0 59 dist[rear] = dist[front]+1; 60 if(try_to_insert(rear)) 61 { 62 fa[rear] = front; 63 rear ++; 64 } 65 } 66 } 67 front ++; 68 } 69 return 0; 70 } 71 72 void print_ans(int u) 73 { 74 vector<vector<int>> nodes; 75 vector<int> tmp; 76 _for(j,0,9) 77 { 78 tmp.push_back(st[u][j]); 79 } 80 nodes.push_back(tmp); 81 while(1) 82 { 83 tmp.clear(); 84 if(fa[u]==0) break; 85 _for(j,0,9) 86 { 87 tmp.push_back(st[fa[u]][j]); 88 } 89 nodes.push_back(tmp); 90 u = fa[u]; 91 } 92 int cnt = 0; 93 int kase = 0; 94 for(int i = nodes.size()-1; i >= 0; i --) 95 { 96 if(!kase) 97 { 98 cout << "Ori:" << endl; 99 kase ++; 100 } 101 else 102 cout << "Step:" << kase++ << endl; 103 cout << nodes[i][0] << " " << nodes[i][1] << " " << nodes[i][2] << endl; 104 cout << nodes[i][3] << " " << nodes[i][4] << " " << nodes[i][5] << endl; 105 cout << nodes[i][6] << " " << nodes[i][7] << " " << nodes[i][8] << endl; 106 cout << endl << endl; 107 } 108 } 109 110 int main() 111 { 112 //freopen("output.txt", "w", stdout); 113 _for(i,0,9) scanf("%d",&st[1][i]); 114 _for(i,0,9) scanf("%d",&goal[i]); 115 int ans = bfs(); 116 if(ans>0) 117 { 118 printf("Step Needed:%d times ",dist[ans]); 119 print_ans(ans); 120 } 121 else printf("-1 "); 122 return 0; 123 }