POJ 1077 Eight， 八数码问题

题目链接：

http://poj.org/problem?id=1077

题目类型： 隐式图搜索

原题：

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

题目大意：

编号为1～8的8个正方形被摆放成3行3列（留有一个格子为空），与空格相邻的编号格子可以移动到空格中。然后要求求出目标状态的方案。

分析与总结：

据说没做过八数码的人生是不完整的。看了lrj的书先做了这道，用到的是所有方法中最朴素，最简单的一种。直接单向bfs+哈希判重。

在poj上可以水过，但在HDU和ZOJ交就不行了。

在做这道题中学到的几样小技巧：

1. 数组直接用memcpy, memcmp对整块内存进行复制或者比较， 速度比用for循环快。

2.用typedef来定义一个新名称可以更加方便。

3.哈希表与编码的应用

还有很多更好更高级的方法，我的人生还待完整……

#include<iostream>
#include<cstring>
#include<cstdio>
#define MAXN 500000
using namespace std;
char input[30];
int state[9], goal[9] = {1,2,3,4,5,6,7,8,0};
int dir[4][2] = {{-1,0},{1,0},{0,-1},{0,1}}; // 上，下，左， 右
char path_dir[5] = "udlr";
int st[MAXN][9];
int father[MAXN], path[MAXN]; // 保存打印路径

const int MAXHASHSIZE = 1000003;
int head[MAXHASHSIZE], next[MAXN];

void init_lookup_table() { memset(head, 0, sizeof(head)); }

typedef int State[9];
int hash(State& s) {
  int v = 0;
  for(int i = 0; i < 9; i++) v = v * 10 + s[i];
  return v % MAXHASHSIZE;

}

int try_to_insert(int s) {
  int h = hash(st[s]);
  int u = head[h];
  while(u) {
    if(memcmp(st[u], st[s], sizeof(st[s])) == 0) return 0;
    u = next[u];
  }
  next[s] = head[h];
  head[h] = s;
  return 1;
}

int bfs(){
    init_lookup_table();
    father[0] = path[0] = -1;
    int front=0, rear=1;
    memcpy(st[0], state, sizeof(state));
    
    while(front < rear){
        int *s = st[front];
       
        if(memcmp(s, goal, sizeof(goal))==0){
            return front;
        }

        int j;
        for(j=0; j<9; ++j) if(!s[j])break; // 找出0的位置
        int x=j/3, y=j%3;     // 转换成行，列
        
        for(int i=0; i<4; ++i){

            int dx = x+dir[i][0]; // 新状态的行，列
            int dy = y+dir[i][1];
            int pos = dx*3+dy;    // 目标的位置

            if(dx>=0 && dx<3 && dy>=0 && dy<3){
                int *newState = st[rear];
                memcpy(newState, s, sizeof(int)*9);
                newState[j] = s[pos];
                newState[pos] = 0;
                if(try_to_insert(rear)){
                    father[rear] = front;  path[rear] = i;
                    rear++;
                }
            }
        } 
        front++;
    }
    return -1;
}

void print_path(int cur){
    if(cur!=0){
        print_path(father[cur]);
        printf("%c", path_dir[path[cur]]);
    }
}

int main(){
    
    while(gets(input)){
        // 转换成状态数组, 'x'用0代替
        for(int pos=0, i=0; i<strlen(input); ++i){
            if(input[i]>='0' && input[i]<='9')
                state[pos++] = input[i]-'0';
            else if(input[i]=='x')
                state[pos++] = 0;
        }
        int ans;
        if((ans=bfs())!=-1){ 
            print_path(ans);
            printf("\n");
        }
    }        
}

—— 生命的意义，在于赋予它意义。


 

   
 

 


   原创
  http://blog.csdn.net/shuangde800
  ，By D_Double (转载请标明)