timus 1015

最新推荐文章于 2026-01-01 19:48:17 发布

原创最新推荐文章于 2026-01-01 19:48:17 发布 · 233 阅读

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#iterator #数据结构 #pair #ini #list #c#

//http://acm.timus.ru/problem.aspx?space=1&num=1015

//这道题思路完全是出自别人的c#代码.我只是翻译成c++而已.

//在翻译的过程中,采用了C++中list和map数据结构.

//Map用于存放字典,即一种排列(key) 对应的 1-30(value).Map有一个好处,就是当重复插入的时候,直接无视.

//list<int>用来存放输出时的顺序

//Group[30]用来存放每个排列所对应的value

#include <iostream>
#include <stdio.h>
#include <string>
#include <map>
#include <sstream>
using namespace std;
#include <list>

long RotationEX(long die, int p0,int p1,int p2,int p3)
{
stringstream ss;
ss<<die;
string s = ss.str();

   char tmp = s[p0];
   s[p0] = s[p1];
   s[p1] = s[p2];
   s[p2] = s[p3];
   s[p3] = tmp;
   return atol(s.c_str());
}

long Rotation(long die, int x, int y, int z)
{
int i;
   for ( i = 0; i < x; i++) die = RotationEX(die, 2, 3, 4, 5);
   for ( i = 0; i < y; i++) die = RotationEX(die, 0, 2, 1, 4);
   for ( i = 0; i < z; i++) die = RotationEX(die, 0, 3, 1, 5);
   return die;
}

int main()
{

long N,i;
scanf("%ld",&N);
int* Group=new int[N];
int left,right,top,forward,bottom,backward;
map<int,int> myMap;
map<int,int>::iterator iter;

int* flag = new int[N];
memset(flag,0,sizeof(int)*N);
//Get The Map
long qs[] =
      {
         123456, 123465, 123546, 123564, 123645, 123654,
         132456, 132465, 132546, 132564, 132645, 132654,
         142356, 142365, 142536, 142563, 142635, 142653,
         152346, 152364, 152436, 152463, 152634, 152643,
         162345, 162354, 162435, 162453, 162534, 162543
       };

   for (int x = 0; x < 4; x++)
     for (int y = 0; y < 4; y++)
       for (int z = 0; z < 4; z++)
         for (short q = 0; q < 30; q++)
           myMap.insert(make_pair(Rotation(qs[q], x, y, z), q));

for (i=0;i<N;i++)
{
  cin>>left>>right>>top>>forward>>bottom>>backward;

  iter = myMap.find(left*100000+right*10000+top*1000+forward*100+bottom*10+backward);
  if (iter!=myMap.end())
   Group[i]= (*iter).second;

}

list<int> sortEx;
list<int> dice[30];

for (i=0;i<N;i++)
{
  if (dice[Group[i]].empty())
  {
   sortEx.push_back(Group[i]);
   dice[Group[i]].push_back(i+1);
  }
  else
   dice[Group[i]].push_back(i+1);
}

printf("%ld/n",sortEx.size());

list<int>::iterator it,dice_it;

for ( it=sortEx.begin() ; it != sortEx.end(); it++ )
{
  for ( dice_it=dice[*it].begin() ; dice_it != dice[*it].end(); dice_it++ )
   cout << *dice_it<< " ";

cout<<endl;
}

return 0;
}