hdu1043 八数码问题

Eight

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17589    Accepted Submission(s): 4795
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
    
    
    
    

 

从来没有接触过这种搜索题，是在看了别人的题解的提示以后知道怎 么做的。看了别人的题解，觉得高明之处在于使用康托展开判重。按照我最初的想法，把每个数字的摆放看作是一种状态嘛，我直接去搜索就可以了，但是还是有很多麻烦，而使用康托展开直接把当前状态hash省去了很多不必要的麻烦 。 值得一提的是这个题目是特殊判断，之前没有注意，算出结果了以为自己是错的。这个题我并不是直接去搜索，而是反向搜索去遍历每一种状态，并记录路径，然后根据输入查表解决，按照康托+BFS直TLE。

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <ctype.h>
#include <queue>
using namespace std;

const int N=400000;//最多的状态也就是9！

struct node{
   int  s[9];
   int loc;
   int status;//康托哈希
   string path;
};

int vis[N];//存储状态的访问情况
node initcur;//初始状态

//康托判重
//阶乘0! 1! 2! 3! 4! 5! 6! 7! 8! 9!
int fact[]={1,1,2,6,24,120,720,5040,40320,362880};

int Cantor(int s[],int n){
    int i,j,sum=0,num;
    for(i=0;i<n;i++){
        num=0;
        for(j=i+1;j<n;j++){
            if(s[i]>s[j])
                num++;
        }
        sum+=num*fact[n-i-1];
    }
    return sum;
}

//方向 上  右 下  左
int dic[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
char index[5]="udlr";
string path;
string allpath[N];

int solve(){
    queue<node> Q;
    int aim=initcur.status;//结束状态的hash
    while(!Q.empty()){
        Q.pop();
    }
    node next;
    node cur={{1,2,3,4,5,6,7,8,0},8,46233,""};
    Q.push(cur);
    while(!Q.empty()){
        cur=Q.front();
        Q.pop();
        int x,y;
        x=cur.loc/3;
        y=cur.loc%3;
        for(int i=0;i<4;i++){
            int dx=x+dic[i][0];
            int dy=y+dic[i][1];
            if(0<=dx&&dx<3&&0<=dy&&dy<3){
                next=cur;
                next.loc=dx*3+dy;
                next.s[cur.loc]=next.s[next.loc];
                next.s[next.loc]=0;
                next.status=Cantor(next.s,9);
                if(!vis[next.status]){
                    vis[next.status]=1;
                    next.path=next.path+index[i];
                    allpath[next.status]=next.path;
                    Q.push(next);
                }
            }
        }
    }
    return 0;
}

string solvepath(string path){
    string realpath="";
    for(int i=path.length()-1;i>=0;i--){
        if(path[i]=='u')
            realpath+='d';
        else if(path[i]=='d'){
            realpath+='u';
        }
        else if(path[i]=='l'){
            realpath+='r';
        }
        else if(path[i]=='r'){
            realpath+='l';
        }
    }
    return realpath;
}



int main(){
char tile[30];
memset(vis,0,sizeof(vis));
solve();
while(gets(tile)!=NULL){
    int i;
    int num=0;
    for(i=0;tile[i];i++){
        if(isspace(tile[i])){
            continue;
        }
        else{
            if(tile[i]=='x'){
                initcur.s[num]=0;
                initcur.loc=num;
                num++;
            }
            else{
                initcur.s[num]=tile[i]-'0';
                num++;
            }
        }
    }
    initcur.status=Cantor(initcur.s,9);
    if(allpath[initcur.status]!=""){
        cout<<solvepath(allpath[initcur.status])<<endl;
    }
    else{
        cout<<"unsolvable"<<endl;
    }

}
return 0;
}