POJ Eight， 八数码有关问题

POJ 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 500000using 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)

查看更多 下一篇