Eight HDU-1043 （bfs）

Eight

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35625    Accepted Submission(s): 9219
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

 

思路：使用一维的string来表示当前局面，下标从0开始。

　　   但是代码超内存，可能是由于无解的情况导致的，待更新......

 #include <iostream>
 #include <queue>
 #include <cstring>
 #include <cstdio>
 #include <string>
 #include <algorithm>
 #include <set>
 #include <map>

 using namespace std;

 map<string, char> mp;    // 存储当前局面和方向
 map<string, string>pre;    // 存储当前局面和上一局面

 int flag = ;

 struct node
 {
     int cur;       // x在string中的下标
     string s;    // 当前局面
 }nod;

 string Swap(string s, int x, int y)
 {
     swap(s[x], s[y]);
     return s;
 }

 void Print(string str)  // 递归打印结果
 {
     if(mp[str] == '#')
         return;
     Print(pre[str]);
     cout << mp[str];
 }

 void bfs()
 {
     queue<node> Q;

     Q.push(nod);
     mp[nod.s] = '#';

     node p,t;
     while(!Q.empty())
     {
         p = Q.front();
         Q.pop();

         if(p.s == "12345678x")
         {
             flag = ;
             Print("12345678x");
         }

         for(int i = ; i < ; ++i)
         {
             if(i == )    // 向左
             {
                 if(p.cur %  != )  // 下标为0,3,6的不能向左移动
                 {
                     t.s = Swap(p.s, p.cur, p.cur-);
                     if(mp.count(t.s) == )
                     {
                         mp[t.s] = 'l';
                         pre[t.s] = p.s;
                         t.cur = p.cur - ;
                         Q.push(t);
                     }

                 }
             }
             else if(i == )    // 向右
             {
                 if(p.cur %  != )  // 下标为2,5,8的不能向右移动
                 {
                     t.s = Swap(p.s, p.cur, p.cur+);
                     if(mp.count(t.s) == )
                     {
                         mp[t.s] = 'r';
                         pre[t.s] = p.s;
                         t.cur = p.cur + ;
                         Q.push(t);
                     }

                 }
             }
             else if(i == )    // 向上
             {
                 if(p.cur > )   // 下标为0,1,2的不能向上移动
                 {
                     t.s = Swap(p.s, p.cur, p.cur-);
                     if(mp.count(t.s) == )
                     {
                         mp[t.s] = 'u';
                         pre[t.s] = p.s;
                         t.cur = p.cur - ;
                         Q.push(t);
                     }

                 }
             }
             else if(i == )    // 向下
             {
                 if(p.cur < )   // 下标为6,7,8的不能向下移动
                 {
                     t.s = Swap(p.s, p.cur, p.cur+);
                     if(mp.count(t.s) == )
                     {
                         mp[t.s] = 'd';
                         pre[t.s] = p.s;
                         t.cur = p.cur + ;
                         Q.push(t);
                     }
                 }
             }

         }
     }
 }

 int main()
 {

     char c;
     string str;
     int k;
     for(int i = ; i < ; ++i)
     {
         cin >> c;
         str += c;
         if(c == 'x')
             k = i;    // 记录x的初始下标
     }

     nod.s = str;
     nod.cur = k;

     bfs();
     if(flag == )
         cout << "unsolvable";
     cout << endl;

     return ;
 }