題目描述：
The 15-puzzle has been around for over 100 years; even if you don't know it by that name, you've seen it. It is constructed with 15 sliding tiles, each with a number from 1 to 15 on it, and all packed into a 4 by 4 frame with one tile missing. Let's call the missing tile 'x'; the object of the puzzle is to arrange the tiles so that they are ordered as:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 x
where the only legal operation is to exchange 'x' with one of the tiles with which it shares an edge. As an example, the following sequence of moves solves a slightly scrambled puzzle:
1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
5 6 7 8 5 6 7 8 5 6 7 8 5 6 7 8
9 x 10 12 9 10 x 12 9 10 11 12 9 10 11 12
13 14 11 15 13 14 11 15 13 14 x 15 13 14 15 x
r-> d-> r->
The letters in the previous row indicate which neighbor of the 'x' tile is swapped with the 'x' tile at each step; legal values are 'r','l','u' and 'd', for right, left, up, and down, respectively.
Not all puzzles can be solved; in 1870, a man named Sam Loyd was famous for distributing an unsolvable version of the puzzle, and
frustrating many people. In fact, all you have to do to make a regular puzzle into an unsolvable one is to swap two tiles (not counting the missing 'x' tile, of course).
In this problem, you will write a program for solving the less well-known 8-puzzle, composed of tiles on a three by three
arrangement.
Input
You will receive, several descriptions of configuration of the 8 puzzle. One description is just a list of the tiles in their initial positions, with the rows listed from top to bottom, and the tiles listed from left to right within a row, where the tiles are represented by numbers 1 to 8, plus 'x'. For example, this puzzle
1 2 3
x 4 6
7 5 8
is described by this list:
1 2 3 x 4 6 7 5 8
Output
You will print to standard output either the word ``unsolvable'', if the puzzle has no solution, or a string consisting entirely of the letters 'r', 'l', 'u' and 'd' that describes a series of moves that produce a solution. The string should include no spaces and start at the beginning of the line. Do not print a blank line between cases.
Sample Input
2 3 4 1 5 x 7 6 8
Sample Output
ullddrurdllurdruldr
解題報告：
1：我這裏的做法是逆向bfs，從12345678x這個狀態開始bfs，到達每一個點的時候將操作存起來。
2：方向數組要和字典數組反着來，最後答案要逆向輸出。可能我這題
代碼：
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N = 4e5+10;
ll Map[3][3], fac[10], dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};
char dic[4] = {'u', 'd', 'l', 'r'};
struct point{
ll num, x, y;
vector<char> V;
}p1, p2;
vector<char> VV[N];
map<ll, ll> M;
queue<point> Q;
inline void init(){
fac[0] = fac[1] = 1;
for(ll i=2; i<10; ++i)fac[i] = fac[i-1]*i;
}
inline ll Hash(){
ll ans = 0;
for(ll i=0; i<9; ++i){
ll cnt = 0;
for(ll j=i+1; j<9; ++j){
if(Map[i/3][i%3] > Map[j/3][j%3])++cnt;
}
ans += cnt*fac[8-i];
}
return ans;
}
inline void toArray(ll x){
ll visit[10];
memset(visit, 0, sizeof(visit));
for(ll i=8; i>=0; --i){
ll cnt = 0, k, idx = 8-i;
for(k=1; k<=9; ++k){
if(!visit[k])++cnt;
if(cnt == x/fac[i]+1)break;
}
Map[idx/3][idx%3] = k, x %= fac[i];
visit[k] = 1;
}
}
void bfs(ll sX, ll sY, ll s){
p1.x = sX, p1.y = sY, p1.num = s, M[s] = 1;
Q.push(p1);
while(!Q.empty()){
p1 = Q.front();
for(ll 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;
toArray(p1.num);
swap(Map[p2.x][p2.y], Map[p1.x][p1.y]);
p2.num = Hash();
if(M[p2.num])continue;
p2.V = p1.V, p2.V.push_back(dic[i]);
VV[p2.num] = p2.V, M[p2.num] = 1;
Q.push(p2);
}
Q.pop();
}
}
int main(){
char x[2];
init(), bfs(2, 2, 0);
while(~scanf("%s", x)){
Map[0][0] = x[0] == 'x' ? 9 : x[0]-'0';
for(ll i=1; i<9; ++i){
scanf("%s", x);
Map[i/3][i%3] = x[0] == 'x' ? 9 : x[0]-'0';
}
ll s = Hash();
if(VV[s].size() == 0 && s != 0)printf("unsolvable");
else {
ll len = VV[s].size();
for(ll i=len-1; i>=0; --i)printf("%c", VV[s][i]);
}
puts("");
}
return 0;
}
