hdu5012 水搜索

题意：

      给你一个正方体的初始状态和末状态，问你是否可以再6步之内转到这个状态，有四种转的方式，如果你面对的是正方向的正前方，那么转的方式就是 顺时针，逆时针，上，下。

思路： 

      水搜索，直接搜就行，深搜广搜随意，mark不mark也随意，因为状态4，深度6那么最多也就4^6，题目没有坑点，细心点就行了，我写的是一个广搜（mark了）。

#include<stdio.h>
#include<queue>

using namespace std;

typedef struct
{
   int now[7];
}NODE;

NODE xin ,tou;
int mark[7][7][7][7][7][7];
int ss[7] ,ee[7];

int BFS()
{
   memset(mark ,0 ,sizeof(mark));
   mark[ss[1]][ss[2]][ss[3]][ss[4]][ss[5]][ss[6]] = 1;
   queue<NODE>q;
   for(int i = 0 ;i <= 6 ;i ++) xin.now[i] = ss[i];
   q.push(xin);
   while(!q.empty())
   {
      tou = q.front();
      q.pop();
      int ok = 1;
      for(int i = 1 ;i <= 6 ;i ++)
      if(tou.now[i] != ee[i]) ok = 0;
      if(ok) return tou.now[0];
      if(tou.now[0] >= 6) continue;

      for(int i = 0 ;i <= 6 ;i ++) xin.now[i] = tou.now[i];
      xin.now[0] ++;
      xin.now[1] = tou.now[3] ,xin.now[3] = tou.now[2] ,xin.now[2] = tou.now[4] ,xin.now[4] = tou.now[1];
      if(!mark[xin.now[1]][xin.now[2]][xin.now[3]][xin.now[4]][xin.now[5]][xin.now[6]] ) q.push(xin);

      for(int i = 0 ;i <= 6 ;i ++) xin.now[i] = tou.now[i];
      xin.now[0] ++;
      xin.now[1] = tou.now[4] ,xin.now[4] = tou.now[2] ,xin.now[2] = tou.now[3] ,xin.now[3] = tou.now[1];
      if(!mark[xin.now[1]][xin.now[2]][xin.now[3]][xin.now[4]][xin.now[5]][xin.now[6]] ) q.push(xin);

      for(int i = 0 ;i <= 6 ;i ++) xin.now[i] = tou.now[i];
      xin.now[0] ++;
      xin.now[1] = tou.now[5] ,xin.now[5] = tou.now[2] ,xin.now[2] = tou.now[6] ,xin.now[6] = tou.now[1];
      if(!mark[xin.now[1]][xin.now[2]][xin.now[3]][xin.now[4]][xin.now[5]][xin.now[6]] ) q.push(xin);      

      for(int i = 0 ;i <= 6 ;i ++) xin.now[i] = tou.now[i];
      xin.now[0] ++;
      xin.now[1] = tou.now[6] ,xin.now[6] = tou.now[2] ,xin.now[2] = tou.now[5] ,xin.now[5] = tou.now[1];
      if(!mark[xin.now[1]][xin.now[2]][xin.now[3]][xin.now[4]][xin.now[5]][xin.now[6]] ) q.push(xin);

   }
   return -1;
}

int main ()
{
   while(~scanf("%d %d %d %d %d %d" ,&ss[1] ,&ss[2] ,&ss[3] ,&ss[4] ,&ss[5] ,&ss[6]))
   {
      for(int i = 1 ;i <= 6 ;i ++)
      scanf("%d" ,&ee[i]);
      ss[0] = ee[0] = 0;
      printf("%d\n" ,BFS());
   }
   return 0;
}