Eight（bfs+全排列的哈希函数）

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 22207		Accepted: 9846		Special Judge

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 a description of a configuration of the 8 puzzle. The 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.

Sample Input

 2  3  4  1  5  x  7  6  8 

Sample Output

ullddrurdllurdruldr

题意：这是一个8数码问题，听着好高端的样子；就是给你一个3*3的矩阵，包括1~8和x;例

1  2  3 

x  4  6 

7  5  8 问最少需要变换x几步成为

1  2  3
4  5  6
7  8  x 的形式；

这题的关键是找到一个哈希函数，使得矩阵形成的排列与一个自然数一一对应，这里采用的是全排列的哈希函数，另一个就是BFS加打印路径了；

#include<stdio.h>
#include<string.h>
#include<queue>
#include<iostream>
using namespace std;

const int maxn = 362881;//根据全排列的哈希函数，n+1个数的排列可以对应n个数的多进制形式，这里九个数对应多进制的最大值为9！-1；
int factorial[9] = {1,1,2,6,24,120,720,5040,40320};
int pow[9] = {100000000,10000000,1000000,100000,10000,1000,100,10,1};
int head,tail;
bool vis[maxn];

struct node
{
    char status;
    int id,num,pre;
}que[maxn];

int hash(int num)//全排列的哈希函数
{
    int a[10],key,i,j,c;
    for(i = 0; i < 9; i++)
    {
        a[i] = num%10;//a数组倒着存的num，所以求逆序数的时候条件是a[j]<a[i];
        num = num/10;
    }
    key = 0;
    for(i = 1; i < 9; i++)
    {
        for(j = 0,c = 0; j < i; j++)
        {
            if(a[j] < a[i])
                c++;
        }
        key += c*factorial[i];
    }
    return key;
}

void change(int num,int a,int b,char status)//a位置上的数和b位置上的数互换；
{
    int n1,n2;
    n1 = num/pow[a]%10;
    n2 = num/pow[b]%10;
    num = num - (n1-n2)*pow[a] + (n1-n2)*pow[b];
    int key = hash(num);
    if(!vis[key])
    {
        vis[key] = true;
        que[tail].num = num;
        que[tail].id = b;
        que[tail].status = status;
        que[tail++].pre = head;
    }
}

//打印路径
void print(int head)
{
    char s[100];
    int c = 0;
    while(que[head].status != 'k')
    {
        s[c++] = que[head].status;
        head = que[head].pre;
    }
    s[c] = '\0';
    for(int i = c-1; i >= 0; i--)
    {
        printf("%c",s[i]);
    }
    printf("\n");
}

int main()
{
    char c;
    int num,id,t;

    num = 0;
    for(int i = 0; i < 9; i++)
    {
        cin>>c;
        if(c == 'x')
        {
            t = 9;
            id = i;
        }
        else t = c-'0';
        num = 10*num+t;
    }
    bool flag = false;
    memset(vis,false,sizeof(vis));

    head = 0;
    tail = 1;
    que[0].id = id;
    que[0].num = num;
    que[0].status = 'k';

    while(head < tail)
    {
        num = que[head].num;
        id = que[head].id;
        if(num == 123456789)
        {
            flag = true;
            break;
        }
        if(id > 2)
            change(num,id,id-3,'u');

        if(id < 6)
            change(num,id,id+3,'d');

        if(id%3 != 0)
            change(num,id,id-1,'l');
        if(id%3 != 2)
            change(num,id,id+1,'r');
        head++;

    }
    if(flag) print(head);
    else printf("unsolvable\n");
    return 0;
}