Eight

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

【題目解析】

這道題目是一道典型的搜尋題目，什麼搜尋？自然是廣搜。面對這一道題，其中最為複雜的就是如何儲存？如何判重？你想到了嗎？
面對如何儲存的這一個問題，我選擇了棧，因為它可以更好的幫助我儲存。至於如何判重，我選擇了先用一個一維陣列儲存，再運用以下方法：

若有5個數； 2 4 5 7 8 
我們如何來進行判重呢?
首先我們需要一個 bool 陣列來標記曾經產生過的值，避免重複產生。
那我們如何產生這樣一個值呢？
這時，需要兩個 for 迴圈來輔助，不光如此，我們還需要一個階層的陣列，以輔助產生。
第一個 for 列舉第一個數 i:1~n（從前往後），第二數列舉 j:i+1~n （從前往後），一旦遇到比第一個數小的就加上 [8-j]的階層如此，我們就創造出了一個數所獨有的自己的編碼。至於為什麼是 8-j 而不是 j，也是為了運算速度著想，不過，也可以用 j。

知道了這個方法，媽媽再也不用擔心我的判重了（-__-），另外一個需要注意的是：如何判斷是否越界？請細細理解下面這段程式碼：

if(push.where<0||push.where>8||(i%2&&push.where/3!=first.where/3))

push.where and first.where為新的位置編號與原來的位置編號。這裡的判重終點是左右走（判斷的行號）。
好啦，下面自己消化程式碼吧 -_-
【程式碼】
狀態：AC

#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
int Jc[15]={1,1,2,6,24,120,720,5040,40320,362880,3628800};
int Fx[4]={-3,-1,3,1};
char Sc[4]={'u','l','d','r'};
struct Queue
{
        int num[15];
        int Dad,F,where;
}a[600000];
Queue begin;
bool map[362885];
bool Dd(int n[])   
{
    for(int i=0;i<8;i++)
        if(n[i]!=i+1)
            return false;
    return true;
}
bool Pc(int n[])
{
    int lj=0;
    for(int i=0;i<9;i++) 
        for(int j=i+1;j<9;j++) 
            if(n[j]<n[i])   
                lj+=Jc[8-i];
    if(map[lj]) return true;
    map[lj]=true;
    return false;
}
int main()
{
    int l=0,r=0;
    char c=0;
    begin.Dad=begin.F=-1;
    for(int i=0;i<9;i++)
    {
        scanf("%c",&c);
        if(c-'0'==-16)scanf("%c",&c);
        if(c!='x')begin.num[i]=c-'0';
        else    begin.num[i]=0,begin.where=i;
    }
    //for(int i=0;i<9;i++)
        //printf("%d ",begin.num[i]);
    a[r++]=begin;
    while(l<r)
    {
            Queue first;
            first=a[l];
            for(int i=0;i<4;i++)
            {
                Queue push=first;
                push.Dad=l,push.F=i,push.where=Fx[i]+first.where;
                if(push.where<0||push.where>8||(i%2&&push.where/3!=first.where/3)) continue;
                    swap(push.num[push.where],push.num[first.where]);
                if(Pc(push.num)) continue;
                    a[r++]=push;
                if(Dd(push.num))
                {
                    int x=r-1;
                    vector<int> vec;
                    while(true)
                    {
                    vec.push_back(a[x].F);
                    x=a[x].Dad;
                    if(x==-1) break;
                    }
                    for(int j=vec.size()-2;j>=0;j--)
                    printf("%c",Sc[vec[j]]);
                    printf("\n");
                    return 0;
                }
            }
            l++;
    }
    printf("NO SOLUTION\n");
    return 0;
}