八数码问题的初次解决（康托展开+bfs）

八数码问题，常常是很多书上的例题，他作为一道经典的搜索题，被很多人誉为 不做过人生不完整，然而，对于这道题，实在是有太多高端的解法，然而我对于A*的理解还颇有偏差，于是选择了最普通的算法去完成这个问题，事实上，对于仅仅是八个数的题目来说，这道题目还是有很多方法可以解决的，下面我就简单的讲讲我做的那一种，而目前能力水平上的局限，希望能够在以后慢慢填补，下一次再用A*把这道题目做出来。

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

题目大意就是将八个数放在3*3的格子里，通过移动’X’的各自使他达到目标状态

解法：这题可以直接用BFS搜索，难度在于怎么标记状态，这里我们可以把X转化成0 在用康托展开，使得每一个数找到其对应的位置，这样就能用一个VIS 来标记。

#include <algorithm>
#include <bitset>
#include <cassert>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <deque>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#define inf 0x3f3f3f3f
#define eps 1e-6
using namespace std;
typedef long long ll;
using namespace std;
const int dx[] = { -1,1,0,0 };
const int dy[] = { 0,0,-1,1 };
char op[] = { 'u','d','l','r' };
typedef int state[9];
struct node {
    state s;
    int d;
    int pre;
    char op;
};
node ei;
vector<node>m;
state goal;
int vis[362880];
int fact[9];
void init() {
    fact[0] = 1;
    int i;
    for (i = 1; i < 9; i++) {
        fact[i] = fact[i - 1] * i;
    }
}
int try_to_insert(node t) {
    int code = 0;
    int i;
    for (i = 0; i<9; i++) {
        int cnt = 0;
        int j;
        for (j = i + 1; j<9; j++)
            if (t.s[j]<t.s[i])cnt++;
        code += fact[8 - i] * cnt;
    }
    if (vis[code])return 0;
    return vis[code] = 1;
}
node temp, cha;
int bfs() {
    queue<node>q;
    ei.d = 0;
    ei.pre = -1;
    ei.op = '#';
    q.push(ei);
    int z;
    while (!q.empty()) {
        temp = q.front();
        m.push_back(temp);
        q.pop();
        if (memcmp(goal, temp.s, sizeof(temp.s) == 0)) {
            return temp.d;
        }
        for (z = 0; z < 9; z++) {
            if (!temp.s[z])
                break;
        }
        int x = z / 3, y = z % 3;
        int d;
        for (d = 0; d < 4; d++) {
            int newx = x + dx[d];
            int newy = y + dy[d];
            int newz = newx * 3 + newy;
            if (newx >= 0 && newx < 3 && newy >= 0 && newy < 3) {
                memcpy(cha.s, temp.s, sizeof(temp.s));
                int t;
                t = cha.s[newz];
                cha.s[z] = t;
                cha.s[newz] = 0;
                if (try_to_insert(cha) == 0)continue;
                cha.d = temp.d + 1;
                cha.pre = m.size() - 1;
                cha.op = op[d];
                q.push(cha);
                if (memcmp(goal, cha.s, sizeof(cha.s)) == 0) {
                    return cha.d;
                }
            }

        }
    }
    return 0;
}
char seq[1234567];
int main() {
    init();
    char a[9];
    for (int i = 0; i < 9; i++) {
        goal[i] = i + 1;
    }
    goal[8] = 0;
    for (int i = 0; i < 9; i++) {
        cin >> a[i];
        if (a[i] == 'x')ei.s[i] = 0;
        else ei.s[i] = a[i] - '0';
    }
    if (bfs() == 0)printf("unsolvable\n");
    else {
        int sn = 1, k;
        k = cha.pre;
        seq[0] = cha.op;
        while (m[k].pre != -1)
        {
            seq[sn++] = m[k].op;
            k = m[k].pre;
        }
        sn = sn - 1;
        int i;
        for (i = sn; i >= 0; i--)
        {
            cout << seq[i] ;
        }
        cout << endl;
    }
    return 0;
}