HDU1043 Eight(八数码,BFS,康托展开,hash)

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

思路

这是一个八数码问题,关于这道题的题意,可以参考Virtual Judge上面的翻译,我们的思路是从初始状态开始搜索,然后搜索完所有的状态,每次搜索的时候记录路径,并且采用康托展开进行判重,因为我们已经预处理好了所有的状态。所以最后应对 O(1) O ( 1 ) 的查询就好了

代码

#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
const int N=4e5+10;
int fac[10]= {0,1,2,6,24,120,720,5040,40320,362880};
string path[N];
int vis[N];
int go[4][2]= {-1,0,1,0,0,-1,0,1}; //上下左右
char dir[]="durl";//和方向变量相反
struct node
{
    int s[9];//序列
    int zero_pos;//0的位置
    int hash;//哈希值
    string path;//路径
};
int cantor_hash(int s[])//康托展开判重
{
    int res=0;
    for(int i=0; i<9; i++)
    {
        int num=0;
        for(int j=i+1; j<9; j++)
            if(s[i]>s[j])
                num++;
        res+=(num*fac[9-i-1]);
    }
    return res;
}
void init()
{
    node now,to;
    mem(vis,0);
    for(int i=0; i<8; i++)
        now.s[i]=i+1;
    now.s[8]=0;
    now.zero_pos=8;
    now.hash=cantor_hash(now.s);
    path[now.hash]="";
    vis[now.hash]=1;
    queueq;
    q.push(now);
    while(!q.empty())
    {
        now=q.front();
        q.pop();
        int x=now.zero_pos/3;
        int y=now.zero_pos%3;
        for(int i=0; i<4; i++)
        {
            int xx=x+go[i][0];
            int yy=y+go[i][1];
            if(xx>=0&&xx<3&&yy>=0&&yy<3)
            {
                to=now;
                to.zero_pos=xx*3+yy;
                to.s[now.zero_pos]=to.s[to.zero_pos];
                to.s[to.zero_pos]=0;
                to.hash=cantor_hash(to.s);
                if(!vis[to.hash])
                {
                    to.path=dir[i]+now.path;
                    path[to.hash]=to.path;
                    q.push(to);
                    vis[to.hash]=1;
                }
            }
        }
    }
}

int main()
{
    init();
    node ans;
    char s[5];
    while(~scanf("%s",s))
    {
        if(s[0]=='x')
        {
            ans.zero_pos=0;
            ans.s[0]=0;
        }
        else
            ans.s[0]=s[0]-'0';
        for(int i=1; i<9; i++)
        {
            scanf("%s",s);
            if(s[0]=='x')
            {
                ans.zero_pos=i;
                ans.s[i]=0;
            }
            else
                ans.s[i]=s[0]-'0';
        }
        ans.hash=cantor_hash(ans.s);
        if(vis[ans.hash])
            cout<else
            puts("unsolvable");
    }
    return 0;
}

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