HDU - 1043 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

题解：一个3*3的矩阵，求最少要怎样移动“x”能使矩阵变成 1 2 3 4 5 6 7 8 x。类似于拼图游戏（讨厌又不会玩。。。）

看了很多大佬的博客才磕磕绊绊。。。。还是会的太少太少了。。。

用bfs倒着搜索，记录每种状态，最后判定一下输出就好

代码：

#include<stdio.h>
#include<string.h>
#include<queue>
#include<map>
#include<string>
#include<algorithm>
using namespace std;
int nextt[4][2]= {0,1,1,0,0,-1,-1,0};
map<int,string>lu;//记录路径
map<int,int>f;//标记作用，防止状态重复
struct node
{
    string t;
    int mp[3][3];//存放矩阵
    int x,y;
};
void dfs()
{
    queue<node>q;
    node st,ed;
    int l=1;
    for(int i=0; i<3; i++)
        for(int j=0; j<3; j++)
            st.mp[i][j]=l++;//初始状态
    st.mp[2][2]=0;//最后一个是X，记为0
    st.x=2;
    st.y=2;
    f[123456780]=1;
    q.push(st);
    while(!q.empty())
    {
        st=q.front();
        q.pop();
        ed=st;
        for(int i=0; i<4; i++)
        {
            int tx=st.x+nextt[i][0];
            int ty=st.y+nextt[i][1];
            if(tx<0||ty<0||tx>=3||ty>=3)
                continue;
            for(int j=0; j<3; j++)
                for(int k=0; k<3; k++)
                    ed.mp[j][k]=st.mp[j][k];
            swap(ed.mp[st.x][st.y],ed.mp[tx][ty]);  //X和别的交换位置
            int sum=1;
            for(int j=0; j<3; j++)
                for(int k=0; k<3; k++)
                    sum=sum*10+ed.mp[j][k];   //计算此时状态
            if(f[sum]==0)
            {
                f[sum]=1;
                ed.x=tx;
                ed.y=ty;
                if(i==0)
                    ed.t=st.t+'l';    //反向存路径
                else if(i==1)
                    ed.t=st.t+'u';
                else if(i==2)
                    ed.t=st.t+'r';
                else
                    ed.t=st.t+'d';
                q.push(ed);
                lu[sum]=ed.t;   //记录此状态下的路径
            }
        }
    }
}
int main()
{
    dfs();
    char s[50];
    while(gets(s))
    {
        int i,l=strlen(s),sum=1;
        for(i=0; i<l; i++)
        {
            if(s[i]>='1'&&s[i]<='9')
                sum=sum*10+s[i]-'0';
            else if(s[i]=='x')
                sum=sum*10;
        }
        if(f[sum]==0)
            printf("unsolvable\n");
        else
        {
            string ss=lu[sum];
            for(i=ss.size()-1; i>=0; i--)//逆序输出
                printf("%c",ss[i]);
            printf("\n");
        }
    }
    return 0;
}