搜索进阶----A - Eight

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, several descriptions of configuration of the 8 puzzle. One 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. Do not print a blank line between cases.
Sample Input
2 3 4 1 5 x 7 6 8
Sample Output
ullddrurdllurdruldr

这道题有好多做法，我用的是反向bfs打表，有时候思维要正逆结合，把x看做是0，那么这个list就可以变成一个数表示，所有的状态就是9！，大概是1e5次个，从123456780 bfs遍历出所有可达的状态，这是不会超的
注意：
1>是多组输入，原先没看到
2>如果输入是12345678x，则什么也不输出

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<queue>
#include<map>
using namespace std;
typedef long long LL;
const int inf = 0x3f3f3f3f;

int a[10];
map<int,string>mp;
typedef struct Node{
    int num;
    int pos;
}Node;

int Pow(int x)
{
    int sum = 1;
    for(int i = 0;i < x;++i)
    {
        sum *= 10;
    }
    return sum;
}

int main()
{
    //cout << sum << endl;
    int x = 123456780;
    queue<Node>que;
    que.push((Node){x,9});
    while(!que.empty())
    {
        Node tmp = que.front();
        que.pop();
        int num = tmp.num;
        int pos = tmp.pos;
        //cout << tmp.num << endl;
        if((pos - 3) >= 1 && (pos - 3) <= 9){
            int p = num / Pow(9 - pos) % 10;
            //cout << pos << " " << num << " " << Pow(12 - pos) << endl;
            int q = num / Pow(12 - pos) % 10;
            //cout << p << " " << q << endl;
            int ans = num - p * Pow(9 - pos) - q * Pow(12 - pos) + p * Pow(12 - pos) + q * Pow(9 - pos);
            //cout << ans << " " << mp[ans].size() << endl;
            if(!mp[ans].size() || mp[ans].size() > (mp[num].size() + 1)){
                mp[ans] = mp[num] + 'd';
                que.push((Node){ans,pos - 3});
            }
        }
        if((pos + 3) >= 1 && (pos + 3) <= 9){
            int p = num / Pow(9 - pos) % 10;
            int q = num / Pow(6 - pos) % 10;
            int ans = num - p * Pow(9 - pos) - q * Pow(6 - pos) + p * Pow(6 - pos) + q * Pow(9 - pos);
            if(!mp[ans].size() || mp[ans].size() > (mp[num].size() + 1)){
                mp[ans] = mp[num] + 'u';
                que.push((Node){ans,pos + 3});
            }
        }
        if((pos - 1) >= 1 && (pos - 1) <= 9 && (pos - 1) % 3 != 0){
            int p = num / Pow(9 - pos) % 10;
            int q = num / Pow(10 - pos) % 10;
            int ans = num - p * Pow(9 - pos) - q * Pow(10 - pos) + p * Pow(10 - pos) + q * Pow(9 - pos);
            if(!mp[ans].size() || mp[ans].size() > (mp[num].size() + 1)){
                mp[ans] = mp[num] + 'r';
                que.push((Node){ans,pos - 1});
            }
        }
        if((pos + 1) >= 1 && (pos + 1) <= 9 && (pos + 1) % 3 != 1){
            int p = num / Pow(9 - pos) % 10;
            int q = num / Pow(8 - pos) % 10;
            int ans = num - p * Pow(9 - pos) - q * Pow(8 - pos) + p * Pow(8 - pos) + q * Pow(9 - pos);
            if(!mp[ans].size() || mp[ans].size() > (mp[num].size() + 1)){
                mp[ans] = mp[num] + 'l';
                que.push((Node){ans,pos + 1});
            }
        }
    }
    char ch;
    while(~scanf(" %c",&ch)){
        int sum = 0;
        a[1] = 1;
        if(ch == 'x'){
            sum = sum * 10;
        }else{
            sum = sum * 10 + ch - '0';
        }
        for(int i = 2;i <= 9;++i)
        {
            char ch;
            scanf(" %c",&ch);
            a[i] = i % 9;
            if(ch == 'x'){
                sum = sum * 10;
            }else{
                sum = sum * 10 + ch - '0';
            }
        }
        string s = mp[sum];
        if(sum == x){
            printf("\n");
            continue;
        }
        if(s.size() == 0){
            printf("unsolvable\n");
            continue;
        }
        int len = s.size();
        for(int i = len - 1;i >= 0;i--)
        {
            printf("%c",s[i]);
        }
        printf("\n");
    }
    return 0;
}