Eight HDU - 1043

​​Eight HDU - 1043 ​​

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.
题意：
由15个滑块构造的，每个滑块（瓷砖由1-15组成）。x表示空格，r，l,d,u表示移动x的移动操作。题目让解决8个滑块构成的方格（个人觉得就是把蒙娜丽莎的微笑谜图还原）
输入：八个字符代表一个排列，把移动成12345678X
输出：如果可以找到解（12345678X）就输出操作，不然就输出“unsolvable”，注意输出的格式。
解题思路：还是和前一个博客一样的思路，首先打表–1.编码 2康托展开 3.记录父节点和操作字符 4.判断移动的合法性

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<stack>
using namespace std;
const int maxn=370000;
int kangtuo[10];
char op[maxn];//记录到来的操作符 
int fa[maxn];//记录父节点 
int vis[maxn];//记录是否可以到达的状态 
char dir[]="urld";//注意因为是打表，反向的所以 上下左右是反的 ！！！
int dx[]={3,-1,1,-3};


int get_code(string s)
{
    int len=s.size();
    int ans=0;
    for(int i=0;i<len;i++)
    {
        int cnt=0;
        for(int j=i+1;j<len;j++)
        if(s[i]-'0'>s[j]-'0') cnt++;
        ans+=cnt*kangtuo[8-i];
    }
//  cout<<"get_code: "<<ans<<endl; 
    return ans;
}
int move(int w,int x)
{
    if(dx[x]==1&&w%3==2) return 0;
    if(dx[x]==-1&&w%3==0) return 0;
    if(dx[x]==-3&&w<3) return 0;
    if(dx[x]==3&&w>5) return 0;
    return 1;
}
void bfs(string s,int u)
{
    memset(vis,0,sizeof(vis)); 
    int tmp=get_code(s);
    vis[tmp]=1;
    op[tmp]='0';
    fa[tmp]=tmp;

    queue<string> q;
    queue<int> wei;

    q.push(s);
    wei.push(u);

    while(!q.empty())
    {
        string str=q.front();
        int w=wei.front();
        q.pop(),wei.pop();
        int str_code=get_code(str);

        for(int i=0;i<4;i++)
        {
            int nx=w+dx[i];
            if(move(w,i))
            {
                string str2=str;
                str2[nx]=str[w];
                str2[w]=str[nx];
            //  cout<<str2<<endl;

                int tt=get_code(str2);

                if(!vis[tt])
                {
                    vis[tt]=1;
                    op[tt]=dir[i];//dlru 
                    fa[tt]=str_code;//前一个
                    q.push(str2);
                    wei.push(nx); 
                }
            }
        }
    }
}
int main()
{
    kangtuo[0]=1;
    for(int i=1;i<9;i++)
    kangtuo[i]=kangtuo[i-1]*i;//cout<<kangtuo[i]<<endl; ;

    string str="123456780";
    bfs(str,8);//打表
    int code_str=get_code(str);
    string s;
    while(getline(cin,s))
    {
        for(int i=0;i<s.size();i++)
        {
            if(s[i]==' ') s.erase(i,1),i--;
            if(s[i]=='x') s[i]='0';
        }
        //cout<<s<<endl; 
        int code_s=get_code(s);
        //cout<<code_s<<endl;
        if(!vis[code_s])
        printf("unsolvable\n") ;
        else
        {
            while(code_s!=code_str)
            {
                printf("%c",op[code_s]);
                code_s=fa[code_s];
                //cout<<op[code_s]<<endl;
            }
            printf("\n");
        }
    }

    return 0;
}