题目描述：
Eight-puzzle, which is also called "Nine grids", comes from an old game.
In this game, you are given a 3 by 3 board and 8 tiles. The tiles are numbered from 1 to 8 and each covers a grid. As you see, there is a blank grid which can be represented as an 'X'. Tiles in grids having a common edge with the blank grid can be moved into that blank grid. This operation leads to an exchange of 'X' with one tile.
We use the symbol 'r' to represent exchanging 'X' with the tile on its right side, and 'l' for the left side, 'u' for the one above it, 'd' for the one below it.
A state of the board can be represented by a string S using the rule showed below.
The problem is to operate an operation list of 'r', 'u', 'l', 'd' to turn the state of the board from state A to state B. You are required to find the result which meets the following constrains:
1. It is of minimum length among all possible solutions.
2. It is the lexicographically smallest one of all solutions of minimum length.
Input
The first line is T (T <= 200), which means the number of test cases of this problem.
The input of each test case consists of two lines with state A occupying the first line and state B on the second line.
It is guaranteed that there is an available solution from state A to B.
Output
For each test case two lines are expected.
The first line is in the format of "Case x: d", in which x is the case number counted from one, d is the minimum length of operation list you need to turn A to B.
S is the operation list meeting the constraints and it should be showed on the second line.
Sample Input
2
12X453786
12345678X
564178X23
7568X4123Sample Output
Case 1: 2
dd
Case 2: 8
urrulldr解题报告：
1：这题做法就不说了，网上的题解肯定写的比我好。主要我说下我在做这题遇到的问题。
2：首先我的想法是和Eight I一样，把每种情况的答案存起来，果断ME了。然后就是考虑牺牲速度来解决内存了。即把每个状态的上一个状态的康托值和操作记录起来。好了记录下来，过了样例。提交还是ME。郁闷了，我就把所有的long long 改为int。好了解决了ME。提交变T了。=.=整个人都不好了，然后一直排查不出来，过了会再看这题，发现我用的是map标记的，汗。其实可以用代码中的now二维数组来标记，为-1就说明还没遍历过，不为-1就continue。
3：至此，此题A了。也可以说是蒟蒻的一大进步吧。
代码：
#include<bits/stdc++.h>
using namespace std;
const int N = 400000+10;
int dir[4][2] = {1, 0, 0, -1, 0, 1, -1, 0}, now[9][N], pre[9][N], fac[10];
char dic[5] = {'d', 'l', 'r', 'u'};
struct point{
int s, num[9], x, y;
}p1, p2;
queue<point> Q;
inline void init(){
fac[0] = fac[1] = 1;
for(int i=2; i<10; ++i)fac[i] = fac[i-1]*i;
memset(now, -1, sizeof(now));
memset(pre, -1, sizeof(pre));
}
inline int Hash(int x[9]){
int ans = 0;
for(int i=0; i<9; ++i){
int cnt = 0;
for(int j=i+1; j<9; ++j)if(x[i] > x[j])++cnt;
ans += cnt*fac[8-i];
}
return ans;
}
void bfs(point p, int pos){
Q.push(p), now[pos][p.s] = 1;
while(!Q.empty()){
p1 = Q.front();
for(int i=0; i<4; ++i){
p2.x = p1.x+dir[i][0];
p2.y = p1.y+dir[i][1];
if(p2.x < 0 || p2.y < 0 || p2.x >= 3 || p2.y >= 3)continue;
memcpy(p2.num, p1.num, sizeof(p1.num));
swap(p2.num[p2.x*3+p2.y], p2.num[p1.x*3+p1.y]);
p2.s = Hash(p2.num);
if(now[pos][p2.s] != -1)continue;
Q.push(p2);
now[pos][p2.s] = i, pre[pos][p2.s] = p1.s;
}
Q.pop();
}
}
void solve(int e, int pos){
string ans = "";
int x = pre[pos][e];
ans += dic[now[pos][e]];
while(x != -1)
ans += dic[now[pos][x]], x = pre[pos][x];
int len = ans.size()-1;
printf("%d\n", len);
for(int i=len-1; i>=0; --i)printf("%c", ans[i]);puts("");
}
int main(){
init();
for(int i=0; i<9; ++i){
int cnt = 0;
for(int j=0; j<9; ++j)
p1.num[j] = i == j ? 9 : ++cnt;
p1.x = i/3, p1.y = i%3, p1.s = Hash(p1.num);
bfs(p1, i);
}
//printf("ok!\n");
int t, cas = 1;
scanf("%d", &t);
while(t--){
char A[15], B[15];
map<char, char> MM;
int pos, cnt = 0;
scanf("%s%s", A, B);
printf("Case %d: ", cas++);
for(int i=0; i<9; ++i){
if(A[i] == 'X')pos = i, MM['9'] = '9';
else MM[A[i]] = '1' + cnt++;
}
for(int i=0; i<9; ++i)
p1.num[i] = B[i] == 'X' ? 9 : MM[B[i]] - '0';
int e = Hash(p1.num);
solve(e, pos);
}
return 0;
}
