POJ1077&&HDU1043(八数码，IDA*+曼哈顿距离)

Eight

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 30127		Accepted: 13108		Special Judge

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 a description of a configuration of the 8 puzzle. The 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.

Sample Input

 2  3  4  1  5  x  7  6  8 

Sample Output

ullddrurdllurdruldr

 //2016.8.25
 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 #include<map>

 using namespace std;

 int a[][], b[], d[], sx, sy, deep;
 bool ok;
 map<int, bool> vis;
 char dir[] = {'u', 'd', 'l', 'r'};
 int dx[] = {-, , , };//分别对应上下左右四个方向
 int dy[] = {, , -, };

 bool solve()//求逆序对判断是否有解，偶数有解，奇数无解
 {
     int cnt = ;
     for(int i = ; i <= ; i++)
       for(int j = ; j < i; j++)
         if(b[i] && b[j]>b[i])
           cnt++;
     return !(cnt%);
 }

 int Astar()//启发函数，假设每个数字可以从别的数字头上跨过去，计算到达自己应该到的位置所需要的步数，即计算该数到达其正确位置的曼哈顿距离，h为各点曼哈顿距离之和
 {
     int h = ;
     for(int i = ; i <= ; i++)
         for(int j = ; j <= ; j++)
             if(a[i][j]!=)
             {
                 int nx = (a[i][j]-)/;
                 int ny = (a[i][j]-)%;
                 h += (abs(i-nx-)+abs(j-ny-));
             }
     return h;
 }

 int toInt()//把矩阵转换为int型数字
 {
     int res = ;
     for(int i = ; i <= ; i++)
       for(int j = ; j <= ; j++)
         res = res*+a[i][j];
     return res;
 }

 void IDAstar(int x, int y, int step)
 {
     if(ok)return ;
     int h = Astar();
     if(!h && toInt()==)//找到答案
     {
         for(int i = ; i < step; i++)
             cout<<dir[d[i]];
         cout<<endl;
         ok = ;
         return ;
     }
     if(step+h>deep)return ;//现实＋理想<现状，则返回，IDA*最重要的剪枝
     int now = toInt();
     if(vis[now])return ;//如果状态已经搜过了，剪枝，避免重复搜索
     vis[now] = true;
     for(int i = ; i < ; i++)
     {
         int nx = x+dx[i];
         int ny = y+dy[i];
         if(nx>=&&nx<=&&ny>=&&ny<=)
         {
             d[step] = i;
             swap(a[x][y], a[nx][ny]);
             IDAstar(nx, ny, step+);
             swap(a[x][y], a[nx][ny]);
             d[step] = ;
         }
     }
     return;
 }

 int main()
 {
     char ch;
     while(cin >> ch)
     {
         ok = false;
         deep = ;
         int cnt = ;
         for(int i = ; i <= ; i++)
         {
             for(int j = ; j <= ; j++)
             {
                 if(i==&&j==);
                 else cin >> ch;
                 if(ch == 'x')
                 {
                     a[i][j] = ;
                     sx = i;
                     sy = j;
                 }else
                   a[i][j] = ch - '';
                 b[cnt++] = a[i][j];
             }
         }
         if(!solve())
         {
             cout<<"unsolvable"<<endl;
             continue;
         }
         while(!ok)
         {
             vis.clear();
             IDAstar(sx, sy, );
             deep++;//一层一层增加搜索的深度
         }
     }

     return ;
 }