EIGHT 八段码

Eight

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17098    Accepted Submission(s): 4702
Special Judge

Problem Description

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
   
   
   
   

 

Source

South Central USA 1998 (Sepcial Judge Module By JGShining)

 

Recommend

JGShining   |   We have carefully selected several similar problems for you:   1044  1072  1180  1401  1195 

 

八段码，搜索入坑第一题。第一个是单广加打表，状态的存储用康拓定理。

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define maxn 400000
#define aim 46234
using namespace std;
int vis[maxn];
char d[4]={'d','u','r','l'};
int dx[4]={-1,1,0,0};
int dy[4]={0,0,-1,1};
struct node{
   int loc;
   int s[9];
   char dir;
   int status;
   int fa; 
}n[maxn];
char path[maxn][40];
int fac[10] = { 1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880 };

int inverse_cantor(int s[9])
{
    int sum=0,num=0;
    for(int i=0;i<9;i++)
    {
        num=0;
        for(int j=i+1;j<9;j++)
        {
            if(s[j]<s[i])
                num++;
        }
        sum+=num*fac[8-i];
    }
    return sum+1;
}

void count_path(node a)
{
    int status=a.status;
    int f=a.fa;
    int num=0;
    path[status][num++]=a.dir;
    while(f)
    {
        path[status][num++]=n[f].dir;
        f=n[f].fa;
    }
}

void bfs()
{
    memset(vis,0,sizeof(vis));
    node next;
    for(int i=0;i<8;i++)
        n[0].s[i]=i+1;
    n[0].s[8]=0;
    n[0].status=aim;
    n[0].loc=8;
    vis[aim]=1;
    int head=0,tail=0;
    while(head<=tail)
    {
        int x=n[head].loc/3;
        int y=n[head].loc%3;
        for(int i=0;i<4;i++)
        {
            int tx=x+dx[i];
            int ty=y+dy[i];
            if(tx<0||tx>2||ty<0||ty>2) continue;
            next=n[head];
            next.dir=d[i];
            next.fa=head;
            next.loc=3*tx+ty;
            next.s[n[head].loc]=next.s[next.loc];
            next.s[next.loc]=0;
            next.status=inverse_cantor(next.s);
            if(vis[next.status]==0)
            {
                vis[next.status]=1;
                count_path(next);
                n[++tail]=next;
            }
        }
        head++;
    }
}

int main()
{
	bfs();
    int sum=0;
    char ch[3];
    node cur;
    while(scanf("%s",ch)!=EOF)
    {
        if(ch[0]=='x')
        {
            cur.s[0]=0;
            cur.loc=0;
        }
        else cur.s[0]=ch[0]-'0';
        for(int i=1;i<9;i++)
        {
            scanf("%s",ch);
            {
                if(ch[0]=='x')
                {
                    cur.s[i]=0;cur.loc=i;
                }
                else cur.s[i]=ch[0]-'0';
            }
        }
            cur.status=inverse_cantor(cur.s);
            if(vis[cur.status]==1)
                printf("%s\n",path[cur.status]);
            else printf("unsolvable\n");
    }
    return 0;
}

第二个是IDA*

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char op[4]={'u','r','d','l'};
int maze[4][4];
char ip[11];
int pos[9][2] = {
    {0,0},{0,1},{0,2},
    {1,0},{1,1},{1,2},
    {2,0},{2,1},{2,2}
};
int count_step[50];
int limit,ok;
int dx[4]={-1,0,1,0};
int dy[4]={0,1,0,-1};

int h(int x,int y)
{
    int num=0;
    for(int i=0;i<3;i++)
    {
        for(int j=0;j<3;j++)
        {
            if(i!=x||j!=y)
            {
                num+=abs(i-pos[maze[i][j]-1][0]);
                num+=abs(j-pos[maze[i][j]-1][1]);
            }
        }
    }
    return num;
}

int judge()
{
    int maze1[10],k=0,sum=0;
    for(int i=0;i<3;i++)
    {
        for(int j=0;j<3;j++)
        {
            if(maze[i][j]!=9)
            {
                maze1[++k]=maze[i][j];
            }
        }
    }
    for(int i=1;i<=8;i++)
    {
        for(int j=1;j<i;j++)
        {
            if(maze1[i]<maze1[j]) sum++;
        }
    }
    if(sum&1) return 1;
    else return 0;
}

int dfs(int x,int y,int step,int pre)
{
    int hn=h(x,y);
    if(hn==0)
    {
        ok=1;
        return step;
    }
    if((hn+step)>limit) return hn+step;
    int minn=1<<29;
    for(int i=0;i<4;i++)
    {
        if(abs(i-pre)==2)continue;
        int tx=x+dx[i];
        int ty=y+dy[i];
        if(tx<0||tx>=3||ty<0||ty>=3) continue;
        count_step[step]=i;
        swap(maze[x][y],maze[tx][ty]);
        int tmp=dfs(tx,ty,step+1,i);
        if(ok) return tmp;
        minn=min(tmp,minn);
        swap(maze[tx][ty],maze[x][y]);
    }
    return minn;
}

void ida(int x,int y)
{
    ok=0;
    memset(count_step,-1,sizeof(count_step));
    while(ok==0&&limit<=30)
    {
        limit=dfs(x,y,0,-111);
    }
    if(ok==0)
    {
        cout<<"unsolvable"<<endl;
        return ;
    }
    else 
    {
        for(int i=0;i<limit;i++)
        {
            printf("%c",op[count_step[i]]);
        }
        cout<<endl;
        return ;
    }
}

int main()
{
    while(cin>>ip[0])
    {
        for(int i=1;i<9;i++)
            cin>>ip[i];
        int x,y,t=0;
        for(int i=0;i<3;i++)
        {
            for(int j=0;j<3;j++)
            {
                if(ip[t]=='x') 
                {
                    maze[i][j]=9;
                    x=i;y=j;
                    t++;
                }
                else maze[i][j]=ip[t++]-'0';
            }
        }
        limit=h(x,y);
        if(limit==0)
        {
            puts("");
            continue;
        }
        if(judge())
        {
            puts("unsolvable");
        }
        else ida(x,y);
    }
}

A*以后写