1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<queue>
  5 #include<stack>
  6 #include<string>
  7 using namespace std;
  8 string str1, str2;
  9 bool vis[370000];
 10 int dx[4] = { 1,0,0,-1 };
 11 int dy[4] = { 0,-1,1,0 };
 12 char cs[4] = { ‘d‘,‘l‘,‘r‘,‘u‘ };
 13 int pre[10][370000];
 14 int op[10][370000];
 15 int jc[12];
 16 int kt(int s) //康托展开  
 17 {
 18     int code = 0;
 19     int st[9];
 20     for (int i = 8; i >= 0; i--, s /= 10)
 21         st[i] = s % 10;
 22     for (int i = 0; i<9; i++)
 23     {
 24         int cnt = 0;
 25         for (int j = i + 1; j<9; j++)
 26             if (st[j]<st[i]) cnt++;
 27         code += jc[8 - i] * cnt;
 28     }
 29     return code;
 30 }
 31 int skt = 0;
 32 int mypow(int x, int y) {
 33     int ans = 1;
 34     while (y) {
 35         if (y & 1)ans *= x;
 36         x *= x;
 37         y /= 2;
 38     }
 39     return ans;
 40 }
 41 void bfs(string str,int x) {
 42     memset(vis, 0, sizeof(vis));
 43     queue<int>pq;
 44     queue<int>pq2;
 45     queue<int>pq3;
 46     while (!pq.empty()) {
 47         pq.pop();
 48     }
 49     while (!pq3.empty()) {
 50         pq3.pop();
 51     }
 52     while (!pq2.empty()) {
 53         pq2.pop();
 54     }
 55     int tmps = 0;
 56     for (int i = 0; i < 9; i++) {
 57         tmps = tmps * 10 + str[i] - ‘0‘;
 58     }
 59     pq.push(tmps);
 60     pq2.push(x);
 61     int kt000 = kt(tmps);
 62     pq3.push(kt000);
 63     vis[kt000] = 1;
 64     while (!pq.empty()) {
 65         int str0 = pq.front(); pq.pop();
 66         //cout << str0 << endl;
 67         int s0 = pq2.front(); pq2.pop();
 68         int kt010 = pq3.front(); pq3.pop();
 69         for (int i = 0; i < 4; i++) {
 70             int x0 = s0 / 3;
 71             int y0 = s0 % 3;
 72             int tx = x0 + dx[i];
 73             int ty = y0 + dy[i];
 74             int s00 = tx * 3 + ty;
 75             if (tx >= 0 && ty >= 0 && ty < 3 && tx < 3) {
 76                 int str00=str0;
 77                 int skt1 = ((str0) / (mypow(10, (8 - s0)))) % 10;
 78                 str00 -= skt1*mypow(10,(8-s0));
 79                 int skt2= ((str0) / (mypow(10, (8 - s00)))) % 10;
 80                 str00 += skt2 * mypow(10,(8-s0));
 81                 str00 -= skt2 * mypow(10, (8 - s00));
 82                 str00 += skt1 * mypow(10, (8 - s0));
 83                 //str00[s00] = str0[s0];
 84                 int kt0 = kt(str00);
 85                 //skt++;
 86             //    cout << skt << endl;
 87             //    cout << kt0 << endl;
 88                 if (!vis[kt0]) {
 89                     vis[kt0] = 1;
 90                     op[x][kt0] = i;
 91                     pre[x][kt0] = kt010;
 92                     pq.push(str00);
 93                     pq2.push(s00);
 94                     pq3.push(kt0);
 95                 }
 96             }
 97         }
 98     }
 99 
100 }
101 int main() {
102     int t;
103     jc[0] = 1;
104     for (int i = 1; i < 9; i++) {
105         jc[i] = jc[i - 1] * i;
106     }
107     int case1 = 1;
108     string str[9];
109     str[0] = "012345678";
110     bfs(str[0], 0);
111 //    cout << "%%%%%\n";
112     str[1] = "102345678";
113     bfs(str[1], 1);
114     str[2] = "120345678";
115     bfs(str[2], 2);
116     str[3] = "123045678";
117     bfs(str[3], 3);
118     str[4] = "123405678";
119     bfs(str[4], 4);
120     str[5] = "123450678";
121     bfs(str[5], 5);
122     str[6] = "123456078";
123     bfs(str[6], 6);
124     str[7] = "123456708";
125     bfs(str[7], 7);
126     str[8] = "123456780";
127     bfs(str[8], 8);
128     scanf("%d", &t);
129     while (t--) {
130         cin >> str1 >> str2;
131         int u;
132         for (int i = 0; i < 9; i++) {
133             if (str1[i] == ‘X‘) {
134                 str1[i] = ‘0‘;
135                 u = i;
136             }
137             if (str2[i] == ‘X‘) {
138                 str2[i] = ‘0‘;
139             }
140         }
141         char hash0[10];
142         for (int i = 0; i < 9; i++) {
143             hash0[str1[i] - ‘0‘] = str[u][i];
144         }
145         string tmp = "";
146         for (int i = 0; i < 9; i++) {
147             tmp += hash0[str2[i] - ‘0‘];
148         }
149         int s1=0, s2=0;
150         for (int i = 0; i < 9; i++) {
151             s1 = s1 * 10 + str[u][i] - ‘0‘;
152         }
153         for (int i = 0; i < 9; i++) {
154             s2 = s2 * 10 + tmp[i] - ‘0‘;
155         }
156         int sta = kt(s1);
157         int en = kt(s2);
158         stack<int>stk;
159         while (!stk.empty())stk.pop();
160         while (sta != en) {
161             stk.push(en);
162             en = pre[u][en];
163         }
164         printf("Case %d: %d\n", case1++, stk.size());
165         while (!stk.empty()) {
166             int sss = stk.top();
167             stk.pop();
168             if (sss != sta) {
169                 printf("%c",cs[op[u][sss]]);
170             }
171         }
172         printf("\n");
173     }
174     return 0;
175 }