八数码

题目描述

有3X3 共9 个格子，其中有一个格子是空的，其他格子填满了1~8 之间不同的数字。通过移动格子可以改变数字和空格的位置，现在给你初状态和末状态，请你给出最少的移动步数。

 

输入

输入文件名为eight.in。

给出6 行，每行有3 个数，每两个数用一个空格空开，每个数在0~8 之间，其中0 表示空格。

前3 行表示初始状态，后3 行表示目标状态。

 

输出

输出文件名为eight.out。

输出一行一个数，表示从初始状态移动到目标状态的最少步数。

如果无解则输出-1。

 

 

样例输入

1 2 3

4 5 6

7 8 0

1 2 3

4 6 8

7 5 0

 

样例输出

4

思路：哈希判重+宽搜

代码：

var

        hx:array[0..1000000]of longint;

        st:array[0..200001]of string;

        a:array[0..200001]of longint;

        mb:string;

        i,j,k,x,y,z,s,ls,m:longint;

procedure c(x:longint);

var

        ch:char;

        b:boolean;

begin

        inc(j);

        st[j]:=st[i];

        ch:=st[j][k];

        st[j][k]:=st[j][x];

        st[j][x]:=ch;

        a[j]:=a[i]+1;

        if st[j]=mb then

        begin

                writeln(a[j]);

                halt;

        end;

        if length(st[j])=8 then st[j]:='0'+st[j];

        val(st[j],ls);

        b:=true;

        z:=ls mod m;

        while hx[z]<>0 do

        begin

                if hx[z]=ls then

                begin

                        b:=false;

                        break;

                end;

                inc(z);

        end;

        if b then hx[z]:=ls

        else dec(j);

end;

begin

        assign(input,'eight.in');

        reset(input);

        assign(output,'eight.out');

        rewrite(output);

        for i:=1 to 9 do

        begin

                read(ls);

                x:=x*10+ls;

        end;

        m:=1000007;  //越大越好

        str(x,st[1]);

        if length(st[1])=8 then st[1]:='0'+st[1];

        x:=0;

        for i:=1 to 9 do

        begin

                read(ls);

                x:=x*10+ls;

        end;

        str(x,mb);

        if length(mb)=8 then mb:='0'+mb;

        i:=0;

        j:=1;

        while i<j do

        begin

                inc(i);

                for k:=1 to length(st[i]) do

                if st[i][k]='0' then break;

                case k of

                        1:

                        begin

                                c(2);

                                c(4);

                        end;

                        2:

                        begin

                                c(1);

                                c(3);

                                c(5);

                        end;

                        3:

                        begin

                                c(2);

                                c(6);

                        end;

                        4:

                        begin

                                c(1);

                                c(5);

                                c(7);

                        end;

                        5:

                        begin

                                c(2);

                                c(4);

                                c(6);

                                c(8);

                        end;

                        6:

                        begin

                                c(3);

                                c(5);

                                c(9);

                        end;

                        7:

                        begin

                                c(4);

                                c(8);

                        end;

                        8:

                        begin

                                c(5);

                                c(7);

                                c(9);

                        end;

                        9:

                        begin

                                c(6);

                                c(8);

                        end;

                end;

        end;

        writeln(-1);

end.