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");
}
}
}
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