kuangbin专题二 搜索进阶 A - Eight

A - Eight

  

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. 

InputYou 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 
OutputYou 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

  看了康托展开还有大佬写的程序，自己写了一个，但是到最后还是输出的与样例输出不同，但是按我这个跑出来的也可以完成，不知道哪里出了错。。。。。

#include
#include
#include
#include
#include
using namespace std ;
typedef long long ll;
typedef unsigned long long ull;
const int _max =5e5;
char s[9];
int dx[]={1,-1,0,0};
int dy[]={0,0,1,-1};
char p[]={'u','d','l','r'};
bool vis[_max];
string ste[_max];
int aim=46234;
int factory[]={1,1,2,6,24,120,720,5040,40320,362880}; 
struct node{
	int a[9];
	int x;
	int ha;
	string step;
};
long contor(int str[]){
	long result = 0 ;
	for(int i = 0; i < 9; i++){
		int counted = 0;
		for(int j = i+1; j < 9; j++){
			if(str[i] > str[j])
			counted++;
		}
		result += counted*factory[8-i];
	}
	return result+1;
}
void bfs(){
memset(vis,false,sizeof(vis));
    node now,next;
    for(int i=0;i<8;i++)now.a[i]=i+1;
    now.a[8]=0;
    now.x=8;
    now.ha=aim;
    now.step="";
    queueq;
    q.push(now);
    ste[aim]="";
    while(!q.empty())
    {
        now=q.front();
        q.pop();
        int x=now.x/3;
        int y=now.x%3;
        for(int i=0;i<4;i++)
        {
            int tx=x+dx[i];
            int ty=y+dy[i];
            if(tx<0||tx>2||ty<0||ty>2)continue;
            next=now;
            next.x=tx*3+ty;
            next.a[now.x]=next.a[next.x];
            next.a[next.x]=0;
            next.ha=contor(next.a);
            if(!vis[next.ha])
            {
                vis[next.ha]=true;
                next.step=p[i]+next.step;
                q.push(next);
                ste[next.ha]=next.step;
            }
        }
    }
}
int main(){
	node temp;
	bfs();
	while(cin>>s[0]){
		if(s[0]=='x'){
			temp.a[0]=0;temp.x=0;
		}
		else temp.a[0]=s[0]-'0';
		for(int i = 1;i < 9; i++){
			cin>>s[i];
			if(s[i]=='x'){
				temp.a[i]=0;
				temp.x=i;
			}
			else temp.a[i]=s[i]-'0';
		}
		temp.ha=contor(temp.a);
		if(vis[temp.ha]){
			cout<