HDU 1043 Eight (A* + HASH + 康托展开)

Eight

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 13956    Accepted Submission(s): 3957 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

 

 

 

这题活生生搞了五天，为此学了A*和康托展开，虽然自己写出来了但还不是非常明白，有句话说“如果你真的懂了那么你就应该自己动手做一个”，看来我还没真的懂。

首先把每次矩阵的格局存结构里，然后对整个矩阵用康托展开来进行哈希，得到一个值，以此来判断此种情况是否出现过或是是否到达了终点，有一个很好的剪枝就是用奇偶逆序数来判断是否可解，因为最终情况的逆序数是0，而每次变换不会改变逆序数的奇偶，所以如果初始情况的逆序数是奇数的话就不可解。

不明白的地方是A*的估值函数为何不用f = g + h，而是要用h和g分别作为两个参数来比较，刚开始的时候我把两个参数的位置弄反了， T了一天。都是泪。

 #include <iostream>
 #include <cmath>
 #include <string>
 #include <queue>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 using    namespace    std;

 const    int    SIZE = ;
 const    int    GOAL = ;
 const    int    HASH[] = {,,,,,,,,};
 const    int    UP_DATE[][] = {{,-},{,},{-,},{,}};
 int    PATH[];
 int    PRE[];
 struct    Node
 {
     int    map[SIZE][SIZE];
     int    x,y;
     int    h,g;
     int    hash;
     bool    operator <(const Node a) const
     {
         return    h != a.h ? h > a.h : g > a.g;
     }
 };

 bool    solve_able(const Node & r);
 bool    check(const int,const int);
 void    cal_hash(Node & r);
 void    cal_h(Node & r);
 void    search(const Node & r);
 void    show(void);
 int    main(void)
 {
     Node    first;
     char    s[];

     while(gets(s))
     {
         int    k = ;
         memset(PRE,-,sizeof(PRE));
         memset(PATH,-,sizeof(PATH));
         for(int i = ;i <= ;i ++)
             for(int j = ;j <= ;j ++)
             {
                 if(s[k] >= '' && s[k] <= '')
                     first.map[i][j] = s[k] - '';
                 else    if(s[k] == 'x')
                 {
                     first.map[i][j] = ;
                     first.x = i;
                     first.y = j;
                 }
                 else
                     j --;
                 k ++;
             }
         if(!solve_able(first))
         {
             printf("unsolvable\n");
             continue;
         }
         cal_hash(first);
         if(first.hash == GOAL)
         {
             puts("");
             continue;
         }
         PATH[first.hash] = -;
         first.g = ;
         cal_h(first);
         search(first);
     }

     return    ;
 }

 bool    solve_able(const Node & r)
 {
     int    sum = ,count = ;
     int    temp[];

     for(int i = ;i <= ;i ++)
         for(int j = ;j <= ;j ++)
         {
             temp[count] = r.map[i][j];
             count ++;
         }
     for(int i = ;i < ;i ++)
         for(int j = i + ;j < ;j ++)
             if(temp[j] < temp[i] && temp[j] && temp[i])
                 sum ++;
     return    !(sum & );
 }

 bool    check(const int    x,const    int y)
 {
     if(x >=  && x <=  && y >=  && y <= )
         return    true;
     return    false;
 }

 void    cal_hash(Node & r)
 {
     int    sum = ,count = ,box;
     int    temp[];

     for(int i = ;i <= ;i ++)
         for(int j = ;j <= ;j ++)
         {
             temp[count] = r.map[i][j];
             count ++;
         }
     for(int i = ;i < ;i ++)
     {
         box = ;
         for(int j = i + ;j < ;j ++)
             if(temp[j] < temp[i])
                 box ++;
         sum += (box * HASH[i]);
     }
     r.hash = sum;
 }

 void    search(Node const & r)
 {
     Node    cur,next;

     priority_queue<Node>    que;
     que.push(r);
     while(!que.empty())
     {
         cur = que.top();
         que.pop();
         for(int i = ;i < ;i ++)
         {
             next = cur;
             next.x = cur.x + UP_DATE[i][];
             next.y = cur.y + UP_DATE[i][];
             if(!check(next.x,next.y))
                 continue;
             swap(next.map[cur.x][cur.y],next.map[next.x][next.y]);
             cal_hash(next);

             if(PATH[next.hash] == -)
             {
                 PATH[next.hash] = i;
                 PRE[next.hash] = cur.hash;
                 next.g ++;
                 cal_h(next);
                 que.push(next);
             }
             if(next.hash == GOAL)
             {
                 show();
                 return    ;
             }
         }
     }

 }

 void    cal_h(Node & r)
 {
     int    ans = ;
     for(int i = ;i <= ;i ++)
         for(int j = ;j <= ;j ++)
             if(r.map[i][j])
                 ans += abs(i - ((r.map[i][j] - ) /  + )) + abs(j - ((r.map[i][j] - ) %  + ));
     r.h = ans;
 }

 void    show(void)
 {
     string    ans;
     int    hash = GOAL;

     ans.clear();
     while(PRE[hash] != -)
     {
         switch(PATH[hash])
         {
             case    :ans += 'l';break;
             case    :ans += 'r';break;
             case    :ans += 'u';break;
             case    :ans += 'd';break;
         }
         hash = PRE[hash];
     }
     for(int i = ans.size() - ;i >= ;i --)
         printf("%c",ans[i]);
     cout << endl;
 }