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HDU 1043 Eight 八數碼

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HDU 1043 Eight 八數碼

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)

第一次接觸A* 寫得太想吐了。。。

#include
#include
#include
#include
#include
#include
#include
#include
#include
#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      //優先隊列第一關鍵字為h,第二關鍵字為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) //判斷狀態是否合法  逆序數為偶數
{
    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; ia[j])
                ans++;
    return (ans&1);
}

int get_hash(node tmp)  //獲得康拓值
{
    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; ia[i])
                sum++;
        ans+=Hash[i]*sum;
    }

    return ans;
}

int get_h(node tmp)  //估價函數  每一個點移動到原來地方的最短路程和   曼哈頓距離
{
    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<q;
    q.push(u);
    memset(vis,-1,sizeof(vis));
    memset(pri,-1,sizeof(pri));
    vis[u.Hash]=-2;
    while(!q.empty())
    {
        //cout<>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<