HDU 1043 Eight A*算法+康托展开

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 1213 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 1213 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

 跟小时候玩的那个小拼图的玩具差不多= =~

但是好恐怖啊QAQ

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<queue>using namespace std;const int MAXN=4e5+10;int ha[9]={1,1,2,6,24,120,720,5040,40320};//hash的权值int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};char d[]="udlr";int vis[MAXN];struct node{    int f[3][3];    int x,y;    int g,h;//优先队列的两个关键字    int num;    bool operator<(const node a)const    {        return h+g>a.h+a.g;    }};struct path{    int pre;    char ch;}p[MAXN];void Print(int x){    if(p[x].pre==-1) return ;    Print(p[x].pre);    printf("%c",p[x].ch);}int get_hash(struct node e)//康托展开，压缩{    int a[9];    int i,j;    int k=0;    int ans=0;    for(i=0;i<=2;i++)    {        for(j=0;j<=2;j++)        {            a[k++]=e.f[i][j];        }    }    for(i=0;i<=8;i++)    {        k=0;        for(j=0;j<=i-1;j++)        {            if(a[j]>a[i])k++;        }        ans+=ha[i]*k;    }    return ans;}int get_h(struct node e){    int i,j;    int ans=0;    for(i=0;i<=2;i++)    {        for(j=0;j<=2;j++)        {            if(e.f[i][j])            ans+=abs(i-(e.f[i][j]-1)/3)+abs(j-(e.f[i][j]-1)%3);        }    }    return ans;}void A_star(struct node e){    memset(vis,0,sizeof(vis));    int i,j,k;    struct node a,b;    for(i=0;i<=8;i++)        a.f[i/3][i%3]=(i+1)%9;    int ans=get_hash(a);    e.num=get_hash(e);    e.g=0;    e.h=get_h(e);    vis[e.num]=1;    p[e.num].pre=-1;    if(e.num==ans)    {        printf("\n");        return ;    }    priority_queue<struct node>q;    q.push(e);    while(!q.empty())    {        e=q.top();        q.pop();        for(i=0;i<=3;i++)        {            int xx=e.x+dir[i][0];            int yy=e.y+dir[i][1];            if(xx>=0&&xx<=2&&yy>=0&&yy<=2)            {                a=e;                swap(a.f[e.x][e.y],a.f[xx][yy]);                k=get_hash(a);                if(vis[k])continue;                vis[k]=1;                a.num=k;                a.x=xx;                a.y=yy;                a.g++;                a.h=get_h(a);                p[k].pre=e.num;                p[k].ch=d[i];                if(k==ans)                {                    Print(k);                    printf("\n");                    return ;                }                q.push(a);            }        }    }}int main(){    char a[30];    while(gets(a))    {        int i,j,k;        struct node e;        int len=strlen(a);        for(i=0,j=0;i<=len-1;i++)        {            if(a[i]==' ')continue;            else if(a[i]=='x')            {                e.f[j/3][j%3]=0;                e.x=j/3;                e.y=j%3;            }            else e.f[j/3][j%3]=a[i]-'0';            j++;        }        for(i=0,k=0;i<=8;i++)        {            if(e.f[i/3][i%3]==0) continue;            for(j=0;j<=i-1;j++)            {                if(e.f[j/3][j%3]==0) continue;                if(e.f[j/3][j%3]>e.f[i/3][i%3])k++;            }        }        if(k&1) printf("unsolvable\n");//如果k为奇数        else A_star(e);    }    return 0;}