HDU 1043 Eight 八数码

Eight

Time Limit : 10000/5000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 15   Accepted Submission(s) : 2

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)

第一次接触A*  写得太想吐了。。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<cmath>
#include<map>
#include<string>
#define inf 1<<30
#define eps 1e-7
#define LD long double
#define LL long long
#define maxn 1000000005
using namespace std;
const int ed=322560;
int Hash[9]= {1,1,2,6,24,120,720,5040,40320};
struct node
{
    int Map[3][3];
    int h,g;
    int x,y;
    int Hash;
    bool operator<(const node n1)const      //优先队列第一关键字为h,第二关键字为g
    {
        return h!=n1.h?h>n1.h:g>n1.g;
    }
    bool cheak()
    {
        if(x>=0&&x<3&&y>=0&&y<3)
            return true ;
        return false ;
    }
} u,v;
int vis[400000];
int pri[400000];
int dir[4][2]= {0,1,0,-1,1,0,-1,0};

bool ok(node tmp) //判断状态是否合法  逆序数为偶数
{
    int a[9],k=0;
    for(int i=0; i<3; i++)
        for(int j=0; j<3; j++)
            a[k++]=tmp.Map[i][j];
    int ans=0;
    for(int i=0; i<k; i++)
        for(int j=i+1; j<k; j++)
            if(a[i]&&a[j]&&a[i]>a[j])
                ans++;
    return (ans&1);
}

int get_hash(node tmp)  //获得康拓值
{
    int ans=0;
    int a[9],k=0;
    for(int i=0; i<3; i++)
        for(int j=0; j<3; j++)
            a[k++]=tmp.Map[i][j];

    for(int i=0; i<k; i++)
    {
        int sum=0;
        for(int j=0; j<i; j++)
            if(a[j]>a[i])
                sum++;
        ans+=Hash[i]*sum;
    }

    return ans;
}

int get_h(node tmp)  //估价函数  每一个点移动到原来地方的最短路程和   曼哈顿距离
{
    int ans;
    for(int i=0; i<3; i++)
        for(int j=0; j<3; j++)
            if(tmp.Map[i][j])
            ans+=abs(i-(tmp.Map[i][j]-1)/3)+abs(j-(tmp.Map[i][j]-1)%3);
    return ans;
}

void print()
{
    string str;
    str.clear();
    int st=ed;
    while(pri[st]!=-1)
    {
        if(vis[st]==0)
        {
            str+='r';
        }
        else if(vis[st]==1)
        {
            str+='l';
        }
        else if(vis[st]==2)
        {
            str+='d';
        }
        else
            str+='u';
        st=pri[st];
    }
    for(int i=str.size()-1;i>=0;i--)
        cout<<str[i];
    cout<<endl;
    return ;
}

void bfs()
{
    priority_queue<node >q;
    q.push(u);
    memset(vis,-1,sizeof(vis));
    memset(pri,-1,sizeof(pri));
    vis[u.Hash]=-2;
    while(!q.empty())
    {
        //cout<<q.size()<<endl;
        u=q.top();
        q.pop();
        for(int i=0; i<4; i++)
        {
            v=u;
            v.x+=dir[i][0];
            v.y+=dir[i][1];
            if(v.cheak())
            {
                swap(v.Map[v.x][v.y],v.Map[u.x][u.y]);
                v.Hash=get_hash(v);
                if(vis[v.Hash]==-1&&!ok(v))
                {
                    v.h=get_h(v);
                    v.g++;
                    vis[v.Hash]=i;
                    pri[v.Hash]=u.Hash;
                    q.push(v);
                }
            }
                if(v.Hash==ed)
                {
                    print();
                    return ;
                }
        }
    }
}

int main()
{
    char ch[100];
    memset(ch,0,sizeof(ch));
    while(cin>>ch[0])
    {
        for(int i=1; i<9; i++)
            cin>>ch[i];
        for(int i=0; i<3; i++)
        {
            for(int j=0; j<3; j++)
            {
                if(ch[i*3+j]=='x')
                {
                    u.Map[i][j]=0;
                    u.x=i;
                    u.y=j;
                    u.Hash=get_hash(u);
                    u.h=get_h(u);
                    u.g=0;
                }
                else u.Map[i][j]=ch[i*3+j]-'0';
            }
        }

        if(u.Hash==ed)
        {
            cout<<endl;
            continue ;
        }

        if(!ok(u))
            bfs();
        else cout<<"unsolvable"<<endl;
    }
}