二阶魔方每次操作可以将任意一面逆时针或者顺时针旋转90°,如将上面逆时针旋转90°操作如下。
魔方上每一面的优美度就是这个面上4个数字的乘积,而魔方的总优美度就是6个面优美度总和。
现在Nero有一个数字魔方,他想知道这个魔方在操作不超过5次的前提下能达到的最大优美度是多少。
魔方展开后每一块的序号如下图:
输入:
2 -3 -2 3 7 -6 -6 -7 9 -5 -9 -3 -2 1 4 -9 -1 -10 -5 -5 -10 -4 8 2
输出:
8281
基本思路:
初始化一个24大小的数组 存放对应的数字
可以映射出对应每个面的数字组合
每一次转换,只需要将相应的数字交换即可(实际实现 通过一个中间数组),6个面总共有6类转换(逆时针通过三次正向转换)
从原始输入开始,BFS宽度优先搜索,没出现一次新的类别,计算当前的优雅度,更新res
face为对应6个面的数字序号 trans对应转换后每个位置对应原先哪个位置
int arr[24];//24小面
int res = 0;
int face[6][4]
=
{
{0,1,2,3},
{4,5,10,11},
{6,7,12,13},
{8,9,14,15},
{16,17,18,19},
{20,21,22,23}};//存放6次面的旋转 每一次旋转后对应位置是原先那个序号的值
int trans[6][24] =
{{0, 1, 11, 5, 4, 16, 12, 6, 2, 9, 10, 17, 13, 7, 3, 15, 14, 8, 18, 19, 20, 21, 22, 23},//BEHIND{9, 15, 2, 3, 1, 5, 6, 7, 8, 19, 0, 11, 12, 13, 14, 18, 16, 17, 4, 10, 22, 20, 23, 21},//LEFT{20, 1, 22, 3, 10, 4, 0, 7, 8, 9, 11, 5, 2, 13, 14, 15, 6, 17, 12, 19, 16, 21, 18, 23},//RIGHT{0, 7, 2, 13, 4, 5, 6, 17, 14, 8, 10, 11, 12, 19, 15, 9, 16, 21, 18, 23, 20, 1, 22, 3},//UP{2, 0, 3, 1, 6, 7, 8, 9, 23, 22, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 5, 4},//DOWN{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 21, 20, 10, 11, 12, 13, 18, 16, 19, 17, 15, 14, 22, 23}};
核心函数 sum()求解和 trans_fun()实现转换
int sum()
{int res = 0;for (int i = 0; i <6; i++)//6个面相加{int tmp = 1;for (int j = 0; j <4; j++)//4个数字相乘{tmp *= arr[face[i][j]];}res += tmp;}return res;}void trans_fun(int i)
{vector tmp;for (int k = 0; k <24; k++)tmp.push_back(arr[k]);for (int j = 0; j <24; j++)arr[j] = tmp[trans[i][j]];}
BFS搜索过程,通过执行三次trans_fun() 实现逆时针转换,易错点 需要通过4次trans_fun()实现递归的返回改回
void dfs(int depth)
{if (depth == 0)return;for (int j = 0; j <6; j++) //顺时针{trans_fun(j); //每操作完一次 相当于新的组合 更新一次resres = max(res, sum());dfs(depth - 1);trans_fun(j);trans_fun(j);res = max(res, sum());dfs(depth - 1);trans_fun(j);}}int main()
{for (int i = 0; i <24; i++)cin >> arr[i];res = sum();dfs(5);cout <