水题，先把1转到顶部，在使左边尽量小，然后，这个骰子就固定了。然后再编码一下。因为顶都为1，所以就不管了，然后就剩下几个6进制编码。如果两个骰子经过这样的处理后，编码相同，则他们相同。
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/*
PROGRAM: $PROGRAM
AUTHOR: Su Jiao
DATE: 2010-3-15
DESCRIPTION:
$DESCRIPTION
*/
#include <iostream>
using std::cin;
using std::cout;
using std::endl;
#include <sstream>
using std::stringstream;
#include <vector>
using std::vector;
#include <string>
using std::string;
#include <stack>
using std::stack;
#include <queue>
using std::queue;
#include <map>
using std::map;
using std::pair;
using std::make_pair;
#include <algorithm>
using std::sort;
#include <cassert>
//using std::assert;
struct Dice
{
static const int SIDE=6;
static const int TURN=4;
int left,//the number of points on the left side of the die,
right,//then on the right side,
top,//on the top,
forward,//on the forward side,
bottom,//on the bottom
backward;//and on the backward side
int id;
void read()
{
cin>>left>>right>>top>>forward>>bottom>>backward;
}
inline
void swap(int& a,int& b,int& c,int& d)
{
int s=a;
a=b;
b=c;
c=d;
d=s;
}
inline
void turnLR()
{
swap(top,backward,bottom,forward);
}
inline
void turnTB()
{
swap(right,forward,left,backward);
}
inline
void turnFB()
{
swap(right,top,left,bottom);
}
void getid()
{
//let top=1
for (int i=0;(top!=1)&&(i<TURN);i++)
turnTB();
for (int i=0;(top!=1)&&(i<TURN);i++)
turnLR();
for (int i=0;(top!=1)&&(i<TURN);i++)
turnFB();
//assert(top==1);
//let left min
int min_left=SIDE;
for (int i=0;i<TURN;i++)
{
turnTB();
if (min_left>left) min_left=left;
}
for (int i=0;(left!=min_left)&&(i<TURN);i++)
turnTB();
id=((((left)*SIDE+right)*SIDE+forward)*SIDE+bottom)*SIDE+backward;
}
};
class Application
{
int N;
vector<Dice> dice;
public:
Application()
{
cin>>N;
dice.resize(N);
for (int i=0;i<N;i++)
dice[i].read();
}
int run()
{
map<int,vector<int> > idToIndex;
map<int,vector<int> >::iterator it;
for (int i=0;i<N;i++)
{
dice[i].getid();
idToIndex[dice[i].id].push_back(i+1);
}
cout<<idToIndex.size()<<endl;
for (int i=0;i<N;i++)
if ((it=idToIndex.find(dice[i].id))!=idToIndex.end())
{
for (int j=0;j<it->second.size();j++)
cout<<it->second[j]<<char((j+1==it->second.size())?’\n’:’ ‘);
idToIndex.erase(it);
}
return 0;
}
};
int main()
{
Application app;
return app.run();
}