HDU 1043 Eight (A* + HASH + 康托展开)

时间：2015-03-25 18:39:49 阅读：281 评论：0 收藏：0 [点我收藏+]

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 = 5;
const    int    GOAL = 46233;
const    int    HASH[] = {40320,5040,720,120,24,6,2,1,1};
const    int    UP_DATE[][2] = {{0,-1},{0,1},{-1,0},{1,0}};
int    PATH[600000];
int    PRE[600000];
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[50];

   while(gets(s))
   {
       int    k = 0;
       memset(PRE,-1,sizeof(PRE));
       memset(PATH,-1,sizeof(PATH));
       for(int i = 1;i <= 3;i ++)
           for(int j = 1;j <= 3;j ++)
           {
               if(s[k] >= ‘1‘ && s[k] <= ‘9‘)
                   first.map[i][j] = s[k] - ‘0‘;
               else    if(s[k] == ‘x‘)
               {
                   first.map[i][j] = 0;
                   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] = -2;
       first.g = 0;
       cal_h(first);
       search(first);
   }

   return    0;
}

bool    solve_able(const Node & r)
{
   int    sum = 0,count = 0;
   int    temp[10];

   for(int i = 1;i <= 3;i ++)
       for(int j = 1;j <= 3;j ++)
       {
           temp[count] = r.map[i][j];
           count ++;
       }
   for(int i = 0;i < 9;i ++)
       for(int j = i + 1;j < 9;j ++)
           if(temp[j] < temp[i] && temp[j] && temp[i])
               sum ++;
   return    !(sum & 1);
}

bool    check(const int    x,const    int y)
{
   if(x >= 1 && x <= 3 && y >= 1 && y <= 3)
       return    true;
   return    false;
}

void    cal_hash(Node & r)
{
   int    sum = 0,count = 0,box;
   int    temp[10];

   for(int i = 1;i <= 3;i ++)
       for(int j = 1;j <= 3;j ++)
       {
           temp[count] = r.map[i][j];
           count ++;
       }
   for(int i = 0;i < 9;i ++)
   {
       box = 0;
       for(int j = i + 1;j < 9;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 = 0;i < 4;i ++)
       {
           next = cur;
           next.x = cur.x + UP_DATE[i][0];
           next.y = cur.y + UP_DATE[i][1];
           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] == -1)
           {
               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 = 0;
   for(int i = 1;i <= 3;i ++)
       for(int j = 1;j <= 3;j ++)
           if(r.map[i][j])
               ans += abs(i - ((r.map[i][j] - 1) / 3 + 1)) + abs(j - ((r.map[i][j] - 1) % 3 + 1));
   r.h = ans;
}

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

   ans.clear();
   while(PRE[hash] != -1)
   {
       switch(PATH[hash])
       {
           case    0:ans += ‘l‘;break;
           case    1:ans += ‘r‘;break;
           case    2:ans += ‘u‘;break;
           case    3:ans += ‘d‘;break;
       }
       hash = PRE[hash];
   }
   for(int i = ans.size() - 1;i >= 0;i --)
       printf("%c",ans[i]);
   cout << endl;
}
 

HDU 1043 Eight (A* + HASH + 康托展开)

原文：http://www.cnblogs.com/xz816111/p/4366110.html

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