Eight
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32621 Accepted Submission(s): 8488
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
真是秃了……为了这道题去学了康托展开，对这道题现在还是晕晕的……也算是收获挺多的。
题意是：把输入的序列转换成 1 2 3 4 5 6 7 8 x 输出路径，如果不行输出 unslovable
思路来自：https://blog.csdn.net/codeswarrior/article/details/80313475
#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
#include <map>
#include <vector>
#include <queue>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
const int maxn =362880;
int fac[9];
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
char step[4]={'l','r','u','d'};//与dir相反
struct node1
{
int fa;//指示父节点
char step;//记录当前节点走向父节点的方向
}node[maxn];
struct node2
{
int aa[9];//当前序列
int son;//康托展开表示的位置
int n;//9的坐标 n为9在aa中的位置
};
void Fac()//阶乘
{
fac[0]=1;
for(int i=1;i<9;i++)
{
fac[i]=fac[i-1]*i;
}
}
int cantor(int *a)//康托展开
{
int ans=0;
for(int i=0;i<9;i++)
{
int cnt=0;
for(int j=i+1;j<=8;j++)
{
if(a[i]>a[j]) cnt++;
}
ans+=cnt*fac[9-i-1];
}
return ans;
}
void bfs()
{
for(int i=0;i<maxn;i++)
node[i].fa=-1;
queue<node2> q;
node2 a,b;
//从终点开始逆向搜索
for(int i=0;i<9;i++)
{
a.aa[i]=i+1;
}
a.son=maxn;
a.n=8;
node[a.son].fa=0;//指向自己，表明终点
q.push(a);
while(!q.empty())
{
a=q.front();
q.pop();
for(int i=0;i<4;i++)
{
b=a;
int fx=b.n/3;
int fy=b.n%3;//对9坐标从一维转二维
fx+=dir[i][0];
fy+=dir[i][1];
if(fx<0||fy<0||fx>2||fy>2) continue;
b.n=3*fx+fy;//二维转一维
// cout<<"***************"<<endl;
// cout<<"b.aa[b.n]: "<<b.aa[b.n]<<endl;
// cout<<"b.aa[a.n]: "<<b.aa[a.n]<<endl;
// cout<<"***************"<<endl;
//位置转换a.n 走向 b.n
int temp=b.aa[b.n];
b.aa[b.n]=b.aa[a.n];
b.aa[a.n]=temp;
b.son=cantor(b.aa);
if(node[b.son].fa==-1)//说明这个排列是第一次出现，即最少步骤
{
node[b.son].fa=a.son;
node[b.son].step=step[i];//为了后面直接输出，这里与"实际走向相反 "注意！！
q.push(b);
}
}
}
}
int main()
{
Fac();
bfs();
char get[30];
int que[9];
while(gets(get)>0)
{
int k=0;
for(int i=0;get[i]!='\0';i++)
{
if(get[i]>='1'&&get[i]<='8')
que[k++]=get[i]-'0';
else if(get[i]=='x')
que[k++]=9;
}
int s=cantor(que);
// cout<<"11: "<<s<<endl;
// cout<<node[s].fa<<endl;
if(node[s].fa==-1) cout<<"unsolvable"<<endl;
else {
while(node[s].fa!=0)
{
cout<<node[s].step;
s=node[s].fa;
}
cout<<endl;
}
}
return 0;
}其他方法：https://blog.csdn.net/I_am_a_winer/article/details/43898993