【HDU 1043】Eight（A*启发式搜索算法）

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  45  6  7  89 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  45  6  7  8     5  6  7  8     5  6  7  8     5  6  7  89  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  xr->            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

题目大意

这题其实和我们经常玩的一个游戏非常的像，看下图：

不过这边所要拼成的目标是这样的;

1 2 3
4 5 6
7 8 x

题目就是给你任意一个中间的转态，问你怎样移动x（x就相当于图中的白块，可以和相邻块交换。）能到达目标转态。其实这就是一个八数码的经典问题。

思路

八数码问题以及各种实现方法和讲解可以见：http://blog.csdn.net/huangxy10/article/details/8034285
对于这题我把大部分思路都写在了代码注释里，因为A*算法在一定程度上比较难以理解，所以我觉得结合代码大家应该更看的懂，我自己也是看了3天的A*算法后才勉强可以做这题的。在这里我先简要讲一讲这题：
首先这题必须要用hash和康托展开来压缩空间，康托展开的原理我用一个例子来讲解：
对于这样一个排列：14032，他在排列中排第几呢？公式是：
1*4!+3*3!+0*2!+1*1!+0*0!
求法是这样的，首先看第一个数字1，在数字1前面（不是这个排列的前面，而是这个排列所出现的数字中，如14032中只有一个0比它小）小于它的数只有一个0，对于它后面的数可以有4！种（4的全排列），所以对于1就是1*4！，同理对于4，0~4比它小的有0,1,2,3，但是1在前面已经用过了，所以只有3个，对于后面就是3！，所以对于4就是3*3！，同理往后推。

但是为什么要用hash呢，其实就是为了判重，因为如果你用普通的方法去判重，那么对于3*3方块，每一个数字都有9种可能0~8，况且你用普通bfs和dfs的vis数组根本无法判这题状态是否重复。但是hash可以很好的解决这个问题，因为hash可以把3*3这些数字全部排列的个数控制在0~9！上，对于每一个转态就有一个唯一的hash值，此时只要开一个一维的vis[hash最大值]数组就可以判重。

其次这题所用的A*算法见：http://blog.csdn.net/huangxy10/article/details/8034285

代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>using namespace std;
const int maxn=4e5+10;int ha[9]={1,1,2,6,24,120,720,5040,40320};
/*这个ha[9]数组是用来计算hash值的，这个数组中的值分别是0!,1!,2!,3!,4!,5!,6!,7!,8!*/
int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
char d[10]="udlr";
int vis[maxn];//判重 struct proc
{int f[3][3];//当前这个转态每个位置上的值 int x,y;int g,h;int hash_num;bool operator < (const proc a)const{return h+g>a.h+a.g;//优先队列保证 出队的是估值函数值较小的。 }
};struct path
{int pre;char ch;
}p[maxn];//用于记录路径 //评估函数，获得评估值
//计算1~8的数字回到原点需要的步数作为评估值，必定小于实际操作数
int get_h(proc a)
{int i,j,ans=0;for(int i=0;i<3;i++){for(int j=0;j<3;j++){if(a.f[i][j]){ans+=abs(i-(a.f[i][j]-1)/3)+abs(j-(a.f[i][j]-1)%3);//曼哈顿距离}}}return ans;
}//康托展开
int get_hash(proc e)
{int a[9],i,j,k=0,ans=0;  for(i=0;i<3;i++) //将数据排成一排，便于计算 {  for(j=0;j<3;j++)  a[k++]=e.f[i][j];  }  for(i=0;i<9;i++)//计算hash值  {  k=0;  for(j=0;j<i;j++)  if(a[j]>a[i])k++; ans+=ha[i]*k;  }  return ans;
}void print(int x)
{if(p[x].pre==-1) return;print(p[x].pre);printf("%c",p[x].ch);
}void A_star(proc a)
{memset(vis,0,sizeof(vis));int end_ans,xx,yy,k;proc vw,vn;for(int i=0;i<9;i++){vw.f[i/3][i%3]=(i+1)%9;//目标转态时各个位置的值}end_ans=get_hash(vw);//目标转态时的hash值a.hash_num=get_hash(a);a.g=0;a.h=get_h(a);vis[a.hash_num]=1;p[a.hash_num].pre=-1;if(a.hash_num==end_ans){printf("\n");return;}priority_queue<proc> q;q.push(a);while(!q.empty()){a=q.top();q.pop();for(int i=0;i<4;i++){xx=a.x+dir[i][0];yy=a.y+dir[i][1];if(xx<0||yy<0||xx>=3||yy>=3) continue;vw=a;swap(vw.f[a.x][a.y],vw.f[xx][yy]);//交换双方的值k=get_hash(vw);if(vis[k]) continue;vis[k]=1;vw.hash_num=k;vw.x=xx;vw.y=yy;vw.g++;vw.h=get_h(vw);p[k].pre=a.hash_num;p[k].ch=d[i];if(k==end_ans)//如果当前状态的hash值等于目标状态的hash值，表示已经到达目标状态{print(k);printf("\n");return;}q.push(vw);}}
}int main()
{char a[30];  while(gets(a))  {  int i,j,k,n;  proc e;  n=strlen(a);  for(i=0,j=0;i<n;i++)  {  if(a[i]==' ')continue;  if(a[i]=='x'){e.f[j/3][j%3]=0;e.x=j/3;e.y=j%3;}  else e.f[j/3][j%3]=a[i]-'0';  j++;  }   for(i=0,k=0;i<9;i++)  //这边涉及到一个逆序数问题，如果开始状态的逆序数为奇数，那么这个状态只能到达逆序数为奇数的转态，同理偶数逆序数的状态也只能到达逆序数为偶数的状态{  if(e.f[i/3][i%3]==0)continue;  for(j=0;j<i;j++)  {  if(e.f[j/3][j%3]==0)continue;  if(e.f[j/3][j%3]>e.f[i/3][i%3])k++;  }  }  if(k&1)printf("unsolvable\n");  else A_star(e);  }  return 0;
}