ZOJ 3736 Pocket Cube 脑补+BFS

       2*2*2的魔方，给出初始状态和最大步数N，问在旋转不超过N次的情况下，最多能还原魔方的几个面。

       这题的关键是脑补出魔方六种操作的方案...最好的办法是把题上的图抄下来，做成一个正方体....然后就各种轻松加愉快了....由于最大的步数仅仅是7，状态的判重都不用写了，每生成一个状态算一下有几个面是ok的，更新一下最大值就ok了...

       #include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <stack>
#include <map>
using namespace std;
typedef long long ll;
int n,m;
int num;
const int ff[6][24]={
{0,21,2,23,4,5,6,1,9,15,10,11,12,3,8,14,16,7,18,13,20,17,22,19},
{0,7,2,13,4,5,6,17,14,8,10,11,12,19,15,9,16,21,18,23,20,1,22,3},
{2,0,3,1,6,7,8,9,23,22,10,11,12,13,14,15,16,17,18,19,20,21,5,4},
{1,3,0,2,23,22,4,5,6,7,10,11,12,13,14,15,16,17,18,19,20,21,9,8},
{0,1,11,5,4,16,12,6,2,9,10,17,13,7,3,15,14,8,18,19,20,21,22,23},
{0,1,8,14,4,3,7,13,17,9,10,2,6,12,16,15,5,11,18,19,20,21,22,23}
};

struct tube
{
int d[24];
int cnt;
tube(){}
// bool operator<(const tube& kk)const
// {
// for (int i=0; i<24; i++)
// if (d[i]!=kk.d[i]) return d[i]<kk.d[i];
// }
// bool operator==(const tube& kk)const
// {
// bool res=true;
// for (int i=0; i<24; i++)
// if (d[i]!=kk.d[i])
// {
// res=false;
// break;
// }
// return res;
// }
// bool operator>(const tube& kk)const
// {
// for (int i=0; i<24; i++)
// if (d[i]!=kk.d[i]) return d[i]>kk.d[i];
// }
};
tube st;
tube trans(tube tmp,int k)
{
tube res;

for (int i=0; i<24; i++)
res.d[i]=tmp.d[ff[k][i]];

return res;
}
int count(tube tmp)
{
int res=0;
if (tmp.d[0]==tmp.d[1] && tmp.d[1]==tmp.d[2] && tmp.d[2]==tmp.d[3]) res++;
if (tmp.d[8]==tmp.d[9] && tmp.d[14]==tmp.d[15] && tmp.d[8]==tmp.d[14]) res++;
if (tmp.d[6]==tmp.d[7] && tmp.d[7]==tmp.d[12] && tmp.d[12]==tmp.d[13]) res++;
if (tmp.d[4]==tmp.d[5] && tmp.d[5]==tmp.d[11] && tmp.d[11]==tmp.d[10]) res++;
if (tmp.d[16]==tmp.d[17] && tmp.d[17]==tmp.d[19] && tmp.d[19]==tmp.d[18]) res++;
if (tmp.d[20]==tmp.d[21] && tmp.d[21]==tmp.d[22] && tmp.d[22]==tmp.d[23]) res++;
return res;
}

int ans;
void bfs()
{
queue<tube> q;
q.push(st);
// map<tube,int> mp;
// mp[st]=1;
ans=count(st);
while(!q.empty())
{
tube now=q.front();
q.pop();
tube tmp;
for (int i=0; i<6; i++)
{
tmp=trans(now,i);
tmp.cnt=now.cnt+1;
// if (mp.find(tmp)!=mp.end())
// {
// mp[tmp]=1;
// ans=max(ans,count(tmp));
// if (tmp.cnt<num) q.push(tmp);
// }
ans=max(ans,count(tmp));
if (tmp.cnt<num) q.push(tmp);
}
}
}

int main()
{
// freopen("in.txt","r",stdin);
while(cin>>num)
{
for (int i=0; i<24; i++)
{
cin>>st.d[i];
}
st.cnt=0;
bfs();
cout<<ans<<endl;
}
return 0;
}