Eight HDU-1043 （bfs）

Posted 2022-06-12 fengzeng666

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......

  1 #include <iostream>
  2 #include <queue>
  3 #include <cstring>
  4 #include <cstdio>
  5 #include <string>
  6 #include <algorithm>
  7 #include <set>
  8 #include <map>
  9 
 10 using namespace std;
 11 
 12 map<string, char> mp;    // 存储当前局面和方向
 13 map<string, string>pre;    // 存储当前局面和上一局面
 14 
 15 int flag = 0;
 16 
 17 struct node
 18 
 19     int cur;       // x在string中的下标
 20     string s;    // 当前局面
 21 nod;
 22 
 23 string Swap(string s, int x, int y)
 24 
 25     swap(s[x], s[y]);
 26     return s;
 27 
 28 
 29 void Print(string str)  // 递归打印结果
 30 
 31     if(mp[str] == ‘#‘)
 32         return;
 33     Print(pre[str]);
 34     cout << mp[str];
 35 
 36 
 37 void bfs()
 38 
 39     queue<node> Q;
 40 
 41     Q.push(nod);
 42     mp[nod.s] = ‘#‘;
 43 
 44     node p,t;
 45     while(!Q.empty())
 46     
 47         p = Q.front();
 48         Q.pop();
 49 
 50         if(p.s == "12345678x")
 51         
 52             flag = 1;
 53             Print("12345678x");
 54         
 55 
 56         for(int i = 0; i < 4; ++i)
 57         
 58             if(i == 3)    // 向左
 59             
 60                 if(p.cur % 3 != 0)  // 下标为0,3,6的不能向左移动
 61                 
 62                     t.s = Swap(p.s, p.cur, p.cur-1);
 63                     if(mp.count(t.s) == 0)
 64                     
 65                         mp[t.s] = ‘l‘;
 66                         pre[t.s] = p.s;
 67                         t.cur = p.cur - 1;
 68                         Q.push(t);
 69                     
 70 
 71                 
 72             
 73             else if(i == 2)    // 向右
 74             
 75                 if(p.cur % 3 != 2)  // 下标为2,5,8的不能向右移动
 76                 
 77                     t.s = Swap(p.s, p.cur, p.cur+1);
 78                     if(mp.count(t.s) == 0)
 79                     
 80                         mp[t.s] = ‘r‘;
 81                         pre[t.s] = p.s;
 82                         t.cur = p.cur + 1;
 83                         Q.push(t);
 84                     
 85 
 86                 
 87             
 88             else if(i == 1)    // 向上
 89             
 90                 if(p.cur > 2)   // 下标为0,1,2的不能向上移动
 91                 
 92                     t.s = Swap(p.s, p.cur, p.cur-3);
 93                     if(mp.count(t.s) == 0)
 94                     
 95                         mp[t.s] = ‘u‘;
 96                         pre[t.s] = p.s;
 97                         t.cur = p.cur - 3;
 98                         Q.push(t);
 99                     
100 
101                 
102             
103             else if(i == 0)    // 向下
104             
105                 if(p.cur < 6)   // 下标为6,7,8的不能向下移动
106                 
107                     t.s = Swap(p.s, p.cur, p.cur+3);
108                     if(mp.count(t.s) == 0)
109                     
110                         mp[t.s] = ‘d‘;
111                         pre[t.s] = p.s;
112                         t.cur = p.cur + 3;
113                         Q.push(t);
114                     
115                 
116             
117 
118         
119     
120 
121 
122 
123 int main()
124 
125 
126     char c;
127     string str;
128     int k;
129     for(int i = 0; i < 9; ++i)
130     
131         cin >> c;
132         str += c;
133         if(c == ‘x‘)
134             k = i;    // 记录x的初始下标
135     
136 
137     nod.s = str;
138     nod.cur = k;
139 
140     bfs();
141     if(flag == 0)
142         cout << "unsolvable";
143     cout << endl;
144 
145     return 0;
146