来自"NOCOW"

[编辑] 不预处理所有状态的做法

#include <iostream>
 
using namespace std;
//表示每个面相邻的按顺时针顺序的面
int d[7][4] = {{0, 0, 0, 0},
               {3, 4, 5, 6},
               {6, 5, 4, 3},
               {6, 2, 4, 1},
               {1, 3, 2, 5},
               {1, 4, 2, 6},
               {5, 2, 3, 1}};
int a[7], tot;
int h[1000000];
int q[1000000], nt[100000], s1[100000];
 
void calc(int n)
{
	bool b[7] = {0};
	int i, j, k, s, t, ss;
	for(i = 1 ; i <= 6 ; i++)
	{
		cin >> a[i];
		if(a[i] == 1) k = i;
	}
	s = 0;
	for(i = 0 ; i <= 3 ; i++)
		s = s * 10 + a[d[k][i]], b[a[d[k][i]]] = true;
	t = s;
	for(i = 1 ; i <= 3 ; i++)
	{
		s = s % 10 * 1000 + s / 10;
		if(t > s) t = s;
	}
 
	for(i = 2 ; i <= 6 ; i++)
		if(b[i]) j = j * 10 + i;
		else k = i;
	j = 100000 + t * 10 + k;
	if(h[j] == 0) h[j] = n, tot++;
	else nt[q[j]] = n;
	q[j] = n;
	s1[n] = j;
}
 
void work()
{
	int n;
	cin >> n;
	for(int i = 1 ; i <= n ; i++)
		calc(i);
	cout << tot << endl;
	for(int i = 1 ; i <= n ; i++)
	if(s1[i] != -1)
	{
		cout << i;
		for(int j = nt[h[s1[i]]] ; j != 0 ; j = nt[j])
		{
			s1[j] = -1;
			cout << ' ' << j;
		}
		cout << endl;
	}
}
 
int main()
{
	work();
	return 0;
}

[编辑] 先预处理出所有状态,看哪些状态是相同的,再读入数据,用链表存答案,再输出.

program cao;
type
  node=record
    v,next:longint;
  end;
 
var
  f:array[1..6,1..6,1..6,1..6,1..6,1..6] of longint;
  head,order,tail:array[0..720] of longint;
  t:array[0..200000] of node;
  a,b,c,d,e,g,h,i,j,k,l,n,m,p,q,total,a1,a2,a3,a4,a5,a6:longint;
  st:set of 0..6;
 
procedure init;
begin
  total:=0;
  for a1:=1 to 6 do
    for a2:=1 to 6 do
      for a3:=1 to 6 do
        for a4:=1 to 6 do
          for a5:=1 to 6 do
            for a6:=1 to 6 do
            begin
              st:=[a1]+[a2]+[a3]+[a4]+[a5]+[a6];
              if (st=[1..6])and(f[a1,a2,a3,a4,a5,a6]=0) then
              begin
                inc(total);
                f[a1,a2,a3,a4,a5,a6]:=total;
                f[a1,a2,a4,a5,a6,a3]:=total;
                f[a1,a2,a5,a6,a3,a4]:=total;
                f[a1,a2,a6,a3,a4,a5]:=total;
                f[a2,a1,a3,a6,a5,a4]:=total;
                f[a2,a1,a4,a3,a6,a5]:=total;
                f[a2,a1,a5,a4,a3,a6]:=total;
                f[a2,a1,a6,a5,a4,a3]:=total;
                f[a3,a5,a2,a4,a1,a6]:=total;
                f[a3,a5,a6,a2,a4,a1]:=total;
                f[a3,a5,a1,a6,a2,a4]:=total;
                f[a3,a5,a4,a1,a6,a2]:=total;
                f[a5,a3,a1,a4,a2,a6]:=total;
                f[a5,a3,a6,a1,a4,a2]:=total;
                f[a5,a3,a2,a6,a1,a4]:=total;
                f[a5,a3,a4,a2,a6,a1]:=total;
                f[a4,a6,a3,a2,a5,a1]:=total;
                f[a4,a6,a1,a3,a2,a5]:=total;
                f[a4,a6,a5,a1,a3,a2]:=total;
                f[a4,a6,a2,a5,a1,a3]:=total;
                f[a6,a4,a5,a2,a3,a1]:=total;
                f[a6,a4,a1,a5,a2,a3]:=total;
                f[a6,a4,a3,a1,a5,a2]:=total;
                f[a6,a4,a2,a3,a1,a5]:=total;
              end;
            end;
end;
 
procedure main;
begin
  p:=0;
  read(n);
  for i:=1 to n do
  begin
    read(a1,a2,a3,a4,a5,a6);
    inc(total);
    t[total].v:=i;
    if head[f[a1,a2,a3,a4,a5,a6]]=0 then
    begin
      head[f[a1,a2,a3,a4,a5,a6]]:=total;
      tail[f[a1,a2,a3,a4,a5,a6]]:=total;
      inc(p);
      order[p]:=total;
    end
    else
    begin
      t[tail[f[a1,a2,a3,a4,a5,a6]]].next:=total;
      tail[f[a1,a2,a3,a4,a5,a6]]:=total;
    end;
  end;
  writeln(p);
  for i:=1 to p do
  begin
    j:=order[i];
    while j<>0 do
    begin
      write(t[j].v,‘ ‘);
      j:=t[j].next;
    end;
    writeln;
  end;
end;
 
begin
  init;
  main;
end.