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 xwhere 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 }
