HDU 5983.Pocket Cube(2016 CCPC 青岛 B)

489


题目


点击显/隐题目

The Pocket Cube, also known as the Mini Cube or the Ice Cube, is the 2 × 2 × 2 equivalence of a Rubik’s Cube.
The cube consists of 8 pieces, all corners.
Each piece is labeled by a three dimensional coordinate (h, k, l) where h, k, l ∈ {0, 1}. Each of the six faces owns four small faces filled with a positive integer.
For each step, you can choose a certain face and turn the face ninety degrees clockwise or counterclockwise.
You should judge that if one can restore the pocket cube in one step. We say a pocket cube has been restored if each face owns four same integers.






The first line of input contains one integer N(N ≤ 30) which is the number of test cases.
For each test case, the first line describes the top face of the pocket cube, which is the common 2 × 2 face of pieces
labelled by (0, 0, 1),(0, 1, 1),(1, 0, 1),(1, 1, 1). Four integers are given corresponding to the above pieces.
The second line describes the front face, the common face of (1, 0, 1),(1, 1, 1),(1, 0, 0),(1, 1, 0). Four integers are
given corresponding to the above pieces.
The third line describes the bottom face, the common face of (1, 0, 0),(1, 1, 0),(0, 0, 0),(0, 1, 0). Four integers are
given corresponding to the above pieces.
The fourth line describes the back face, the common face of (0, 0, 0),(0, 1, 0),(0, 0, 1),(0, 1, 1). Four integers are
given corresponding to the above pieces.
The fifth line describes the left face, the common face of (0, 0, 0),(0, 0, 1),(1, 0, 0),(1, 0, 1). Four integers are given
corresponding to the above pieces.
The six line describes the right face, the common face of (0, 1, 1),(0, 1, 0),(1, 1, 1),(1, 1, 0). Four integers are given
corresponding to the above pieces.
In other words, each test case contains 24 integers a, b, c to x. You can flat the surface to get the surface development
as follows.


+ - + - + - + - + - + - +
| q | r | a | b | u | v |
+ - + - + - + - + - + - +
| s | t | c | d | w | x |
+ - + - + - + - + - + - +
| e | f |
+ - + - +
| g | h |
+ - + - +
| i | j |
+ - + - +
| k | l |
+ - + - +
| m | n |
+ - + - +
| o | p |
+ - + - +





For each test case, output YES if can be restored in one step, otherwise output NO.






4
1 1 1 1
2 2 2 2
3 3 3 3
4 4 4 4
5 5 5 5
6 6 6 6
6 6 6 6
1 1 1 1
2 2 2 2
3 3 3 3
5 5 5 5
4 4 4 4
1 4 1 4
2 1 2 1
3 2 3 2
4 3 4 3
5 5 5 5
6 6 6 6
1 3 1 3
2 4 2 4
3 1 3 1
4 2 4 2
5 5 5 5
6 6 6 6






YES
YES
YES
NO




题解


题意很容易理解,就是对一个二阶魔方,判断其是否能在一步内还原.
如果会还原三阶魔方,应该很容易就能理解.
经过分析,其实只有7种情况
1种是初始情况就是还原的,另外对于每一面分别向两个方向转一次.

比较稳妥的写法是直接把转后的位置顺序写出来,然后按照这个顺序查询是否存在不合法的情况.
(一定要注意不要写错!!!)

另外,需要额外注意,必须要满足6种颜色每种颜色4个



代码


点击显/隐
`cpp Pocket Cube https://github.com/OhYee/sourcecode/tree/master/ACM 代码备份
/*//
#define debug
#include
//*/
#include
#include
#include
#include
using namespace std;

int cube[24];

int list[7][24] = {
{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23},
{0,5,2,7,4,9,6,11,8,13,10,15,12,1,14,3,16,17,18,19,22,20,23,21},
{0,13,2,15,4,1,6,3,8,5,10,7,12,9,14,11,16,17,18,19,21,23,20,22},
{20,21,2,3,4,5,6,7,8,9,16,17,12,13,14,15,0,1,18,19,10,11,22,23},
{16,17,2,3,4,5,6,7,8,9,20,21,12,13,14,15,10,11,18,19,0,1,22,23},
{2,0,3,1,22,20,6,7,8,9,10,11,12,13,19,17,16,4,18,5,14,21,15,23},
{1,3,0,2,17,19,6,7,8,9,10,11,12,13,20,22,16,15,18,14,4,21,5,23}
};

int check[24]={16,17,0,1,20,21,18,19,2,3,22,23,4,5,6,7,8,9,10,11,12,13,14,15};

map m;
void solve() {
m.clear();
int pos = 0;
int num[24];
memset(num, 0, sizeof(num));

for (int i = 0; i < 6; i++){
for (int j = 0; j < 4; j++) {
int t;
cin >> t;
cube[i * 4 + j] = t;
if (!m.count(t))
m.insert(make_pair(t, pos++));
num[m[t]]++;
}
}

// cout<<"Cube:"<// for(int i=0;i<24;i++)
// cout<// cout<
//必须是6种颜色,并且每种4个

bool ok = true;
for (int i = 0; i < 6 && ok; i++)
if (num[i] != 4)
ok = false;

if (ok) {
ok = false;

for (int i = 0; i < 7 && !ok; i++) {
//分别按照7种情况判断是否能够还原

// cout << "Case" << i << endl;
// for(int j=0;j<6;j++)
// printf("%3d",cube[list[i][check[j]]]);
// cout<// for(int j=6;j<12;j++)
// printf("%3d",cube[list[i][check[j]]]);
// cout<// for(int j=0;j<6;j++)
// printf(" %3d%3dn",cube[list[i][check[12+j2]]],cube[list[i][check[12+j2+1]]]);
// cout<

bool flag = true;
for (int j = 0; j < 6 && flag; j++) {
if (cube[list[i][4  j + 0]] != cube[list[i][4  j + 1]] ||
cube[list[i][4  j + 0]] != cube[list[i][4  j + 2]] ||
cube[list[i][4  j + 0]] != cube[list[i][4  j + 3]])
flag = false;
}
if (flag)
ok = true;
}
}
cout << (ok ? "YES" : "NO") << endl;
}

int main() {
#ifdef debug
freopen("in.txt", "r", stdin);
int START = clock();
#endif

int T;
cin >> T;
while (T--)
solve();

#ifdef debug
printf("Time:%.3f s.n", double(clock() - START) / CLOCKSPERSEC);
#endif
return 0;
}
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