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
经典到我都不用描述大家应该都知道是什么的题目,因为是多组样例，所以我们要反向搜索，即从开始阶段拓展到不能拓展位置，然后记录其中的操作，最后再把所有的结果存起来，输入对应的结果就输出对应的操作就行了，说起来是很简单，的但是其中操作确实并不容易啊，因为特别容易超内存，所以我们要把每个序列都哈希成一个固定的值，所以用到了Cantor展开，然后因为我们不能每个结果都存储到这一步的操作序列（也是超内存）所以我们要记录每个节点的父亲节点和父亲节点到该节点的操作（只有一个）最后我们向上遍历就行了。
代码如下：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<queue>
#include<string>
#include<algorithm>
using namespace std;
const int MAX = 362880;
pair<int,char> p[MAX];//first记录父亲节点，second记录从父亲节点到当前节点的变化。
bool book[MAX];//判断这个节点是否出现过
int Fac[11];//记录阶乘
void init(){
Fac[0] = 1;
for(int i=1;i<=10;++i){
Fac[i] = Fac[i-1]*i;
}
memset(book,0,sizeof(book));
}
//哈希成一个固定的值
int Cantor(string str){
int res = 0,len = (int)str.length();
for(int i=0;i<len;++i){
int Count = 0;
for(int j=i+1;j<len;++j){
if(str[j] < str[i])
Count++;
}
res += Count*Fac[len-i-1];
}
return res;
}
const int xx[] = {0,1,0,-1};
const int yy[] = {1,0,-1,0};
const char ch[] = {'l','u','r','d'};
class Node{
public:
char x[4][4];
int px,py;
Node();
void change(int tx,int ty){
swap(x[tx][ty],x[px][py]);
px = tx,py = ty;
}
};
//把数码棋盘转为字符串
string tostring(char x[][4]){
string res = "";
for(int i=1;i<=3;++i){
for(int j=1;j<=3;++j){
res += x[i][j];
}
}
return res;
}
//求出其位置
int tovalue(Node v){
return Cantor(tostring(v.x));
}
void BFS(Node Begin){
queue<Node> que;
int value = tovalue(Begin);
if(!book[value]){
book[value] = true;
p[0].first = 0;
que.push(Begin);
}
while(!que.empty()){
Node v = que.front();que.pop();
for(int i=0;i<4;++i){
int tx = v.px + xx[i];
int ty = v.py + yy[i];
if(tx < 1 || tx > 3 || ty <1 || ty > 3)
continue;
Node nex = v;
nex.change(tx,ty);
value = tovalue(nex);
if(!book[value]){
book[value] = true;
p[value].first = tovalue(v);
p[value].second = ch[i];
que.push(nex);
}
}
}
}
int main(void){
init();
Node Begin;
for(int i=1;i<=3;++i){
for(int j=1;j<=3;++j){
Begin.x[i][j] = (char)((i-1)*3+j + '0');
}
}
Begin.px = Begin.py = 3;
BFS(Begin);
string str;
while(getline(cin,str)){
Node v;
int Px = 1,Py = 1;
for(int i=0;i<str.length();++i){
if(str[i] != ' '){
if(Py == 4){
Py = 1;
Px++;
}
if(str[i] == 'x'){
v.x[Px][Py] = '9';
v.px = Px;
v.py = Py;
}
else
v.x[Px][Py] = str[i];
Py++;
}
}
if(!book[tovalue(v)]){
cout << "unsolvable" << endl;
}
else{
string res = "";
int k = tovalue(v);//从这个节点向上搜索。
while(k != 0){
res += p[k].second;
k = p[k].first;
}
cout << res << endl;
}
}
return 0;
}
Node::Node(){
px = py = 0;
memset(x,0,sizeof(x));
}