【 POJ - 1077 】 H - Eight （八数码问题） 【 BFS + hash判重 】
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
思路：
用二维数组记录下每种情况的状态图，总情况为9！= 362880。八数码问题最主要是判重，开始用的set判重，但是超时了，然后改为hash表判重
八数码问题也可以参考一下这篇（set判重）：八数码
代码：
#include<iostream>
#include<set>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;
typedef int state[9];
const int maxn=362900; //情况数
const int hashsize=1000003;
int head[hashsize],nexth[hashsize]; //表头，后继
state st[maxn],goal={1,2,3,4,5,6,7,8,0}; //st状态数组，goal目标九宫格
int last[maxn]; //记录上一步
int dx[]={-1,1,0,0}; //四个方向
int dy[]={0,0,1,-1};
char op[maxn];
set<int> s;
char dire(int x,int y) //确定方向
{
if(x==1 && y==0) return 'd';
else if(x==-1 && y==0) return 'u';
else if(x==0 && y==1) return 'r';
else if(x==0 && y==-1)return 'l';
}
int myhash(state &s)
{
int num=0;
for(int i=0;i<9;i++) num = num*10+s[i]; //将状态图转为9位数
return num%hashsize;
}
int try_insert(int rear) //尝试插入
{
int h=myhash(st[rear]);
int u=head[h]; //h所在表的头u
while(u)
{
if(memcmp(st[rear],st[u],sizeof(st[u]))==0) return 0; //不能插
u=nexth[u]; //后继
}
nexth[rear]=head[h]; //头插
head[h]=rear;
return 1;
}
int bfs()
{
fill(head,head+hashsize,0);
int front=1,rear=2;
while(front<rear)
{
state &f = st[front]; 队头状态
if(memcmp(goal,f,sizeof(f))==0) return front; //与目标同，返回
int z=0;
for(;z<9;z++) if(!f[z]) break; //在状态数组中找到0的位置
int nowx=z/3,nowy=z%3; //0的行列位置
for(int i=0;i<4;i++)
{
int newx=nowx+dx[i]; //下一步的x，y
int newy=nowy+dy[i];
int newz=newx*3+newy; //确定在状态数组中的位置
if(newx>=0 && newx<3 && newy>=0 && newy<3) //在范围内
{
state &r=st[rear]; //队尾的位置
memcpy(r,f,sizeof(f)); //先把上一步的状态复制给下一步
r[newz]=f[z]; //然后交换两个位置，实现移动
r[z]=f[newz];
op[rear]=dire(dx[i],dy[i]); //记录该步的方向
last[rear]=front; //记录该步的上一步
if(try_insert(rear)) rear++; //插入成功，队尾往后一步
}
}
front++; //队头可以走的下一步都记录下来了，front++相当于出队
}
return -1; //找不到
}
int main()
{
char c;
for(int i=0;i<9;i++)
{
cin>>c;
if(c=='x') st[1][i]=0; //将x看作0
else st[1][i]=c-'0';
}
int ans=bfs();
if(ans==-1) cout<<"unsolvable"<<endl;
else
{
string ss; //记录步骤
while(ans!=1)
{
ss += op[ans];
ans=last[ans];
}
reverse(ss.begin(),ss.end()); //记录的步骤是反的，要翻转
printf("%s\n",ss.c_str());
}
return 0;
}