poj1077

【题意】

经典的八数码问题。

描述了一个3*3的格子，每个格子里有一个数，还有一个空位，

每次操作可以将空位跟周围的格子里的数交换，问最少多少次操作可以使方格变为

1 2 3

4 5 6

7 8 x

有解打出最少的操作序列

无解打unsolvable

【输入】

一行，表示格子里的数

【输出】

有解打出最少的操作序列

无解打unsolvable

启发式搜索，IDA*算法

估价函数为每个点到自己位置的哈密顿距离之和

program poj1077;const  zl:array [1..4,1..2] of longint=((0,1),(1,0),(-1,0),(0,-1));  change:array [1..4] of char=('d','r','u','l');type  status=array [0..4,0..4] of longint;var  i,j,k,deep,x,y:longint;  square:status;  temp:char;  yes:boolean;  solution:array [0..101] of longint;procedure swap (var a,b:longint);var  i:longint;begin  i:=a;  a:=b;  b:=i;end;function h (now:status):longint;var  i,j,k:longint;begin  k:=0;  for i:=1 to 3 do    for j:=1 to 3 do      if now[i,j]<>9 then      k:=k+abs((now[i,j]-1) div 3 +1-i)+abs((now[i,j]-1) mod 3 +1-j);  h:=k;end;procedure dfs (x,y,g,lasth,last:longint);var  i,j,k,nx,ny:longint;begin  k:=h(square);  if k+g>deep then exit;  if k=0 then    begin      yes:=true;      exit;    end;  for i:=1 to 4 do    if i+last<>5 then    begin      nx:=x+zl[i,1];      ny:=y+zl[i,2];      if (nx>3)or(nx<1)or(ny>3)or(ny<1) then continue;      swap(square[x,y],square[nx,ny]);      solution[g+1]:=i;      dfs(nx,ny,g+1,k,i);      swap(square[x,y],square[nx,ny]);      if yes then exit;    end;end;function idastar(deep:longint):boolean;begin  yes:=false;  dfs(x,y,0,deep,0);  idastar:=yes;end;begin  for i:=1 to 3 do    for j:=1 to 3 do      begin        read(temp);        if temp='x' then          begin            square[i,j]:=9;            x:=i;            y:=j;          end                    else          square[i,j]:=ord(temp)-48;        read(temp);      end;  deep:=h(square);  while not idastar(deep) do    inc(deep);  if deep>100 then    begin      writeln('unsolvable');      exit;    end;  for i:=1 to deep do    write(change[solution[i]]);  writeln;end.