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 a description of a configuration of the 8 puzzle. The 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.
Sample Input
2 3 4 1 5 x 7 6 8
Sample Output
ullddrurdllurdruldr
1 #include<iostream>
2 #include<cstring>
3 #include<cstdio>
4 #define MAXN 500000
5 using namespace std;
6 char input[30];
7 int state[9], goal[9] = {1,2,3,4,5,6,7,8,0};
8 int dir[4][2] = {{-1,0},{1,0},{0,-1},{0,1}}; // 上，下，左， 右
9 char path_dir[5] = "udlr";
10 int st[MAXN][9];
11 int father[MAXN], path[MAXN]; // 保存打印路径
12
13 const int MAXHASHSIZE = 1000003;
14 int head[MAXHASHSIZE], next[MAXN];
15
16 void init_lookup_table() { memset(head, 0, sizeof(head)); }
17
18 typedef int State[9];
19 int hash(State& s) {
20 int v = 0;
21 for(int i = 0; i < 9; i++) v = v * 10 + s[i];
22 return v % MAXHASHSIZE;
23
24 }
25
26 int try_to_insert(int s) {
27 int h = hash(st[s]);
28 int u = head[h];
29 while(u) {
30 if(memcmp(st[u], st[s], sizeof(st[s])) == 0) return 0;
31 u = next[u];
32 }
33 next[s] = head[h];
34 head[h] = s;
35 return 1;
36 }
37
38 int bfs(){
39 init_lookup_table();
40 father[0] = path[0] = -1;
41 int front=0, rear=1;
42 memcpy(st[0], state, sizeof(state));
43
44 while(front < rear){
45 int *s = st[front];
46
47 if(memcmp(s, goal, sizeof(goal))==0){
48 return front;
49 }
50
51 int j;
52 for(j=0; j<9; ++j) if(!s[j])break; // 找出0的位置
53 int x=j/3, y=j%3; // 转换成行，列
54
55 for(int i=0; i<4; ++i){
56
57 int dx = x+dir[i][0]; // 新状态的行，列
58 int dy = y+dir[i][1];
59 int pos = dx*3+dy; // 目标的位置
60
61 if(dx>=0 && dx<3 && dy>=0 && dy<3){
62 int *newState = st[rear];
63 memcpy(newState, s, sizeof(int)*9);
64 newState[j] = s[pos];
65 newState[pos] = 0;
66 if(try_to_insert(rear)){
67 father[rear] = front; path[rear] = i;
68 rear++;
69 }
70 }
71 }
72 front++;
73 }
74 return -1;
75 }
76
77 void print_path(int cur){
78 if(cur!=0){
79 print_path(father[cur]);
80 printf("%c", path_dir[path[cur]]);
81 }
82 }
83
84 int main(){
85
86 while(gets(input)){
87 // 转换成状态数组, ‘x‘用0代替
88 for(int pos=0, i=0; i<strlen(input); ++i){
89 if(input[i]>=‘0‘ && input[i]<=‘9‘)
90 state[pos++] = input[i]-‘0‘;
91 else if(input[i]==‘x‘)
92 state[pos++] = 0;
93 }
94 int ans;
95 if((ans=bfs())!=-1){
96 print_path(ans);
97 printf("\n");
98 }
99 }
100 }