hdu 1043 pku poj 1077 Eight (BFS + 康拓展开)

http://acm.hdu.edu.cn/showproblem.php?pid=1043

http://poj.org/problem?id=1077

Eight

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9173    Accepted Submission(s): 2473 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

 

思路：

八数码问题，我喜欢跟康拓展开联系在一起；

这里解释下康拓展开:

{1,2,3,4,...,n}表示1,2,3,...,n的排列如 {1,2,3} 按从小到大排列一共6个。123 132 213 231 312 321 。

代表的数字 1 2 3 4 5 6 也就是把10进制数与一个排列对应起来。

他们间的对应关系可由康托展开来找到。

如我想知道321是{1,2,3}中第几个大的数可以这样考虑 ：

第一位是3，当第一位的数小于3时，那排列数小于321 如 123、 213 ，小于3的数有1、2 。所以有2*2!个。再看小于第二位2的：小于2的数只有一个就是1 ，

所以有1*1!=1 所以小于321的{1,2,3}排列数有2*2!+1*1!=5个。所以321是第6个大的数。 2*2!+1*1!+0*0!就是康托展开。

再举个例子：1324是{1,2,3,4}排列数中第几个大的数：第一位是1小于1的数没有，是0个 0*3! 第二位是3小于3的数有1和2，但1已经在第一位了，

所以只有一个数2 1*2! 。第三位是2小于2的数是1，但1在第一位，所以有0个数 0*1! ，所以比1324小的排列有0*3!+1*2!+0*1!=2个，1324是第三个大数。

参照本人博客:http://www.cnblogs.com/crazyapple/p/3213725.html 

 

先DFS从：

1 2 3

4 5 6  --->  到所有状态 都事先搜索出来 ，然后用 输入的状态 去中间 回溯查找 需要得到的状态 即可

7 8 x

AC代码：

（提示POJ请用G++提交，用C++在POJ会超时的，我想了原因，可能是STL容器在C++中的耗时远大于G++）

#include <iostream>
#include <stdio.h>
#include <queue>
#include <string.h>

using namespace std;
#define N 363000

struct Nod
{
    int b[10];
    int x,y,pos;
}nd1,nd2;

int fac[] = {1,1,2,6,24,120,720,5040,40320,362880};  //康拓展开用到的数组
 //康托展开：
int cantor(int* a, int k)
{
     int i, j, tmp, num = 0;
     for (i = 0; i < k; i++) {
         tmp = 0;
         for (j = i + 1; j < k; j++)
             if (a[j] < a[i])
                 tmp++;
         num += fac[k - i - 1] * tmp;
     }
     return num;
}

int mark[N],pre[N];
char dir[N];
int cx[]={0,0,1,-1};
int cy[]={-1,1,0,0};

void exchange(int *a,int x,int y)
{
    int temp=a[x];
    a[x]=a[y];
    a[y]=temp;
}

void bfs(int *b,int x,int y)
{
    queue<Nod> q;
    memset(mark,0,sizeof(mark));
    memset(pre,-1,sizeof(pre));
    nd1.x=x;
    nd1.y=y;
    memcpy(nd1.b,b,sizeof(int)*10);
    int i,temp;
    temp = cantor(b,9);
    mark[temp] = 1;
    nd1.pos = temp;
    q.push(nd1);
    while(!q.empty())
    {
        nd2 = q.front();
        q.pop();
        for(i=0;i<4;i++)
        {
            nd1.x = nd2.x + cx[i];
            nd1.y = nd2.y + cy[i];
            if(nd1.x>=0&&nd1.x<3&&nd1.y>=0&&nd1.y<3)
            {
                memcpy(nd1.b,nd2.b,sizeof(int)*10);
                exchange(nd1.b,nd1.x*3+nd1.y,nd2.x*3+nd2.y);
                temp = cantor(nd1.b,9);
                nd1.pos = temp;
                if(mark[temp]==1)  continue;
                mark[temp] = 1;
                pre[temp] = nd2.pos;
                if(cx[i]==0)
                {
                    if(cy[i]==1)   dir[temp] = 'l';   //因为是从目标状态开始搜索的，最后需要倒序输出，然后方向也应该是相反输出
                    else    dir[temp] = 'r';    //dir[temp] = 'l' 的反方向
                }
                if(cy[i]==0)
                {
                    if(cx[i]==1)   dir[temp] = 'u';  //同上
                    else    dir[temp] = 'd';  //同上
                }
                q.push(nd1);
            }
        }
    }
}

int main()
{
    char str[5];
    int a[10],b[10]={1,2,3,4,5,6,7,8,9};  //末状态，从末状态开始搜索
    bfs(b,2,2);  //预先搜索所有状态
    while(~scanf("%s",str))
    {
        if(str[0]=='x')  a[0]=9;
        else a[0]=str[0]-'0';
        int i;
        for(i=1;i<9;i++)
        {
            scanf("%s",str);
            if(str[0]=='x')   a[i]=9;
            else    a[i]=str[0]-'0';
        }
        int temp = cantor(a,9);  //得到起始状态，然后回溯到末状态
        if(!mark[temp])  //所有状态里面若没有起始状态，则为不可达
        {
            printf("unsolvable\n");
            continue;
        }
        while(temp)  //回溯过程
        {
            printf("%c",dir[temp]);
            temp = pre[temp];
        }
        putchar(10);
    }
    return 0;
}