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[];
     int x,y,pos;
 }nd1,nd2;

 int fac[] = {,,,,,,,,,};  //康拓展开用到的数组
  //康托展开：
 int cantor(int* a, int k)
 {
      int i, j, tmp, num = ;
      for (i = ; i < k; i++) {
          tmp = ;
          for (j = i + ; j < k; j++)
              if (a[j] < a[i])
                  tmp++;
          num += fac[k - i - ] * tmp;
      }
      return num;
 }

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

 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,,sizeof(mark));
     memset(pre,-,sizeof(pre));
     nd1.x=x;
     nd1.y=y;
     memcpy(nd1.b,b,sizeof(int)*);
     int i,temp;
     temp = cantor(b,);
     mark[temp] = ;
     nd1.pos = temp;
     q.push(nd1);
     while(!q.empty())
     {
         nd2 = q.front();
         q.pop();
         for(i=;i<;i++)
         {
             nd1.x = nd2.x + cx[i];
             nd1.y = nd2.y + cy[i];
             if(nd1.x>=&&nd1.x<&&nd1.y>=&&nd1.y<)
             {
                 memcpy(nd1.b,nd2.b,sizeof(int)*);
                 exchange(nd1.b,nd1.x*+nd1.y,nd2.x*+nd2.y);
                 temp = cantor(nd1.b,);
                 nd1.pos = temp;
                 if(mark[temp]==)  continue;
                 mark[temp] = ;
                 pre[temp] = nd2.pos;
                 if(cx[i]==)
                 {
                     if(cy[i]==)   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]==)
                 {
                     if(cx[i]==)   dir[temp] = 'u';  //同上
                     else    dir[temp] = 'd';  //同上
                 }
                 q.push(nd1);
             }
         }
     }
 }

 int main()
 {
     char str[];
     int a[],b[]={,,,,,,,,};  //末状态，从末状态开始搜索
     bfs(b,,);  //预先搜索所有状态
     while(~scanf("%s",str))
     {
         if(str[]=='x')  a[]=;
         else a[]=str[]-'';
         int i;
         for(i=;i<;i++)
         {
             scanf("%s",str);
             if(str[]=='x')   a[i]=;
             else    a[i]=str[]-'';
         }
         int temp = cantor(a,);  //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();
     }
     return ;
 }