八码数——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.

InputYou 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
OutputYou 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

 

题意：

给你一个3*3的九方格，每个格子中放入‘1’~‘8’和‘x'中的一个，并且每个格子中的字符不相同；

1 2 3
x 4 6
7 5 8

上面的格子序列可以通过输入 12346758来表示。

其中，’能和与它相邻的上下左右四个格子相交换；交换方式用'u','d','l','r'表示。问能不能通过有限次的交换，使之变为

 1 2 3

 4 5 6

 7 8 x

即 12345678x

若能输出每一步的路径，若不能输出``unsolvable''；

#include <iostream>
#include<queue>
#include<vector>
#include<algorithm>
#include<cstring>
using namespace std;
int fac[]={1,1,2,6,24,120,720,5040,40320,362880};
const int maxn=370000;
bool vis[maxn];
vector<char>Path[maxn];//若用string来保存，则会MLE
int dxy[4][2]={{-1,0},{1,0},{0,-1},{0,1}};//u,d,l,r;
char index[5]="durl";//要与上面相反，因为是反向搜索
int aim=46234;//123456780的hash值
struct node
{
    int loc;//0的位置
    int state;//哈希值
    int s[9];
    vector<char>path;
};
int cantor(int *s)//康拓展开，求hash值
{
    int ans=0;
    for(int i=0;i<9;i++)
    {
        int num=0;
        for(int j=i+1;j<9;j++)
        {
            if(s[j]<s[i])num++;
        }
        ans+=(num*fac[9-i-1]);
    }
    return ans+1;
}
void bfs()
{
    memset(vis,0,sizeof(vis));
    node cur,nex;
    for(int i=0;i<8;i++)cur.s[i]=i+1;
    cur.s[8]=0;
    cur.loc=8;
    cur.state=aim;
    cur.path.clear();
    queue<node>q;
    q.push(cur);
    vis[aim]=1;
    Path[aim].clear();
    while(!q.empty())
    {
        cur=q.front();
        q.pop();
        int x=cur.loc/3,y=cur.loc%3;
        for(int i=0;i<4;i++)
        {
            int dx=x+dxy[i][0],dy=y+dxy[i][1];
            if(dx<0||dx>2||dy<0||dy>2)continue;
            nex=cur;
            nex.loc=dx*3+dy;
            nex.s[cur.loc]=nex.s[nex.loc];//先交换
            nex.s[nex.loc]=0;//再赋值
            nex.state=cantor(nex.s);
            if(!vis[nex.state])
            {
                vis[nex.state]=1;
                nex.path.insert(nex.path.begin(),index[i]);//插入到第一个位置
                q.push(nex);
                Path[nex.state]=nex.path;
            }

        }
    }
}
int main()
{
    char ch;
    node cur;
    vector<char>v;
    vector<char>::iterator it;
    bfs();
    while(cin>>ch)
    {
        if(ch=='x'){cur.s[0]=0;cur.loc=0;}
        else cur.s[0]=ch-'0';
        for(int i=1;i<9;i++)
        {
            cin>>ch;
            if(ch=='x')
            {
                cur.s[i]=0;
                cur.loc=i;
            }
            else cur.s[i]=ch-'0';
        }
        cur.state=cantor(cur.s);
        if(vis[cur.state])
        {
            v=Path[cur.state];
            for(it=v.begin();it!=v.end();it++)
                cout<<*it;
            cout<<endl;

        }
        else cout<<"unsolvable"<<endl;
    }
    return 0;
}