八数码块

时间：2015-07-26 20:30:06 阅读：106 评论：0 收藏：0 [点我收藏+]

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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 }

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八数码块

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原文地址：http://www.cnblogs.com/demodemo/p/4678422.html

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