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 a description of a configuration of the 8 puzzle. The 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.
Sample Input
2 3 4 1 5 x 7 6 8 Sample Output
ullddrurdllurdruldr 題意：問從現在這個狀态回到完整狀态的最小步數，以及路徑。
思路：其實想法還是一個bfs，每個狀态用 123456780 這種9位數來記錄，不過這個9為數太大了，沒辦法開數組，是以我當時就用map來寫。
方法：①.好像是zoj那個八數位問題是有多組資料，是以當時直接從完整狀态（123456780）反向搜，到達所有初始狀态,然後直接用map<int,string>記錄路徑就好。不過這種方法在poj1077就逾時了。原因很簡單嘛，STL極其耗時。
②.我們知道耗時的地方主要是STL，是以就要想辦法避免STL,那麼怎樣對狀态（123456780）進行記錄呢，當然是用哈希了，在這采取康拓展開式（推導不會QAQ）。
康拓展開：（引用 i-Curve的部落格）
定義：康托展開是一個全排列到一個自然數的雙射，常用于建構哈希表時的空間壓縮。 康托展開的實質是計算目前排列在所有由小到大全排列中的名次，是以是可逆的。
原理介紹： X = A[0] * (n-1)! + A[1] * (n-2)! + … + A[n-1] * 0!
A[i] 指的是位于位置i後面的數小于A[i]值的個數,後面乘的就是後面還有多少個數的階乘
說明 ：這個算出來的數康拖展開值，是在所有排列次序 - 1的值，是以X+1即為在全排列中的次序
eg ：在（1，2，3，4，5）5個數的排列組合中，計算 34152的康托展開值。
帶入上面的公式
X = 2 * 4! + 2 * 3! + 0 * 2! + 1 * 1! + 0 * 0!
=>X = 61
康拓展開的構造函數：
//傳回數組a中當下順序的康拖映射
int cantor(int *a,int n)
{
int ans=0;
for(int i=0;i<n;i++)
{
int x=0;int c=1,m=1;//c記錄後面的階乘
for(int j=i+1;j<n;j++)
{
if(a[j]<a[i])x++;
m*=c;c++;
}
ans+=x*m;
}
return ans;
}
逆康拖展開
static const int FAC[] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880}; // 階乘
//康托展開逆運算
void decantor(int x, int n)
{
vector<int> v; // 存放目前可選數
vector<int> a; // 所求排列組合
for(int i=1;i<=n;i++)
v.push_back(i);
for(int i=m;i>=1;i--)
{
int r = x % FAC[i-1];
int t = x / FAC[i-1];
x = r;
sort(v.begin(),v.end());// 從小到大排序
a.push_back(v[t]); // 剩餘數裡第t+1個數為目前位
v.erase(v.begin()+t); // 移除選做目前位的數
}
}
八數位AC代碼（POJ1077）
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
bool book[3000010];
int dir[4][2]= {0,1, 1,0, 0,-1, -1,0};
char F[]= {"rdlu"};
struct node
{
int x,y,h,f,step;
int a[3][3];
} eg[3000010];
int cal(int k) //計算康拓展開值
{
int sum=0;
for(int i=0; i<9; i++)
{
int ans=0; //後面比這個數小的個數
int x=i/3; //對太懶了，直接用的二維數組
int y=i-3*x;
int c=1,m=1; //算階乘
for(int j=i+1; j<9; j++)//計算
{
int tx=j/3;
int ty=j-3*tx;
if(eg[k].a[x][y]>eg[k].a[tx][ty])//計數
ans++;
m*=c; //階乘
c++;
}
sum+=ans*m;//計數
}
return sum;
}
void print(int h)
{
char str[10010];
int n=0;
while(h!=-1)
{
int f=eg[h].f;
str[n++]=F[f];
h=eg[h].h;
}
for(int i=n-2; i>=0; i--) //第一個位置沒有方向的
printf("%c",str[i]);
printf("\n");
}
void bfs()
{
int s=0,e=1;
while(s<e)
{
int x=eg[s].x;
int y=eg[s].y;
for(int i=0; i<4; i++)
{
int tx=x+dir[i][0];
int ty=y+dir[i][1];
if(tx<0||ty<0||tx>=3||ty>=3)
continue;
eg[e]=eg[s];
swap(eg[e].a[x][y],eg[e].a[tx][ty]); //直接換0的位置
int sum=cal(e); //計算康拓展開數值
if(book[sum])continue;//重複
book[sum]=1;
eg[e].step=eg[s].step+1;
eg[e].h=s; //記錄頭節點，友善輸出路徑
eg[e].f=i; //記錄方向
eg[e].x=tx,eg[e].y=ty;
if(sum==46233) //提前算好結束條件下的康拓展開值
{
print(e);
return;
}
e++;
}
s++;
}
printf("unsolvable\n");
}
int main()
{
char str[2];
int a=1;
for(int i=0; i<3; i++)
for(int j=0; j<3; j++)
{
scanf("%s",str);
if(str[0]=='x')
{
eg[0].x=i,eg[0].y=j;
eg[0].a[i][j]=0;
}
else
{
eg[0].a[i][j]=(str[0]-'0');
}
}
eg[0].step=0,eg[0].h=-1;
eg[0].f=0;
int x=cal(0);
book[x]=1;
bfs();
return 0;
}
![Acwing 204. 表達整數的奇怪方式（擴充中國剩餘定理）Acwing 204. 表達整數的奇怪方式（擴充中國剩餘定理）[圖]](data:image/gif;base64,R0lGODlhAQABAIAAAP///wAAACwAAAAAAQABAAACAkQBADs=)