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HDU 1043 Eight 八数码
2015-11-21 00:55:19 来源: 作者: 【 】 浏览:1
Tags:HDU 1043 Eight 八数码

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Eight

Time Limit : 10000/5000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 15 Accepted Submission(s) : 2
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

Source South Central USA 1998 (Sepcial Judge Module By JGShining)

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第一次接触A* 写得太想吐了。。。

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#include
  
   
#include
   
     #include
    
      #include
     
       #include
      
        #include
       
         #include
        
          #include
          #include
          
            #define inf 1<<30 #define eps 1e-7 #define LD long double #define LL long long #define maxn 1000000005 using namespace std; const int ed=322560; int Hash[9]= {1,1,2,6,24,120,720,5040,40320}; struct node { int Map[3][3]; int h,g; int x,y; int Hash; bool operator<(const node n1)const //优先队列第一关键字为h,第二关键字为g { return h!=n1.h?h>n1.h:g>n1.g; } bool cheak() { if(x>=0&&x<3&&y>=0&&y<3) return true ; return false ; } } u,v; int vis[400000]; int pri[400000]; int dir[4][2]= {0,1,0,-1,1,0,-1,0}; bool ok(node tmp) //判断状态是否合法 逆序数为偶数 { int a[9],k=0; for(int i=0; i<3; i++) for(int j=0; j<3; j++) a[k++]=tmp.Map[i][j]; int ans=0; for(int i=0; i
           
            a[j]) ans++; return (ans&1); } int get_hash(node tmp) //获得康拓值 { int ans=0; int a[9],k=0; for(int i=0; i<3; i++) for(int j=0; j<3; j++) a[k++]=tmp.Map[i][j]; for(int i=0; i
            
             a[i]) sum++; ans+=Hash[i]*sum; } return ans; } int get_h(node tmp) //估价函数 每一个点移动到原来地方的最短路程和 曼哈顿距离 { int ans; for(int i=0; i<3; i++) for(int j=0; j<3; j++) if(tmp.Map[i][j]) ans+=abs(i-(tmp.Map[i][j]-1)/3)+abs(j-(tmp.Map[i][j]-1)%3); return ans; } void print() { string str; str.clear(); int st=ed; while(pri[st]!=-1) { if(vis[st]==0) { str+='r'; } else if(vis[st]==1) { str+='l'; } else if(vis[st]==2) { str+='d'; } else str+='u'; st=pri[st]; } for(int i=str.size()-1;i>=0;i--) cout<
             
              q; q.push(u); memset(vis,-1,sizeof(vis)); memset(pri,-1,sizeof(pri)); vis[u.Hash]=-2; while(!q.empty()) { //cout<
              
               >ch[0]) { for(int i=1; i<9; i++) cin>>ch[i]; for(int i=0; i<3; i++) { for(int j=0; j<3; j++) { if(ch[i*3+j]=='x') { u.Map[i][j]=0; u.x=i; u.y=j; u.Hash=get_hash(u); u.h=get_h(u); u.g=0; } else u.Map[i][j]=ch[i*3+j]-'0'; } } if(u.Hash==ed) { cout<
               
                

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