八数码块

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 }

View Code