POJ 1077 Eight（BFS）

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

  类似于拼图游戏，九格拼图怎样才能按顺序拼出来。

  刚看这道题时，我并没有觉得它很难，甚至比之前简单直接搜索不就可以了。后来发现这题目不是这么简单，复杂度真是高了去了。搜的题解也是由多种做法，我目前只看了一种，后几种以后慢慢补上。就这一种方法我也有点不理解，STL中queue的复杂度这么高么，都得手写队列。还有为什么要用康拓展开实现100%的哈希，缩小了所占内存？不哈希我也觉得不是很复杂。先挖好坑，以后慢慢补。我得试试用queue和不哈希能不能挂了，目测应该会挂。多看几遍，希望能理解透。

#include #include #include #include using namespace std; struct Node { int state; int s[9]; int location,pre; char path; }node[1000010]; bool vis[1000010]; int f[9]={1,1,2,6,24,120,720,5040,40320}; int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}}; char op[4]={'u','d','l','r'}; int head,tail; int cantor(int s[]) { int i,j; int num,sum=0; for(i=0;i<9;i++) { num=0; for(j=0;js[i]) num++; sum+=(num*f[i]); } return sum+1; } int bfs(Node now) { int i,j; memset(vis,0,sizeof(vis)); head=tail=0; now.state=cantor(now.s); vis[now.state]=1; node[tail++]=now;//放入队尾，从队尾放元素 Node temp; while(head!=tail)//取出队首元素进行搜索，将符合条件的下一步搜索再放入队列中 { now=node[head++];//取出当前队首元素 if(now.state==1)//123456789的哈希值为1 return head-1; //system("pause"); //cout<<"HEAD:"<=0&&xx<=2&&yy>=0&&yy<=2) { temp=now; temp.s[3*x+y]=temp.s[3*xx+yy];//交换位置 temp.s[3*xx+yy]=9; temp.state=cantor(temp.s); if(!vis[temp.state]) { vis[temp.state]=1; temp.location=xx*3+yy; temp.path=op[i];//保存操作 temp.pre=head-1; node[tail++]=temp; } } } } return -1; } void Print(int i)//压栈，输出路径 { if(i==0) return; Print(node[i].pre); cout<