K - Pocket Cube
Time Limit:10000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit Status Practice HDU 4801
Description
Pocket Cube is a 3-D combination puzzle. It is a 2 × 2 × 2 cube, which means it is constructed by 8 mini-cubes. For a combination of 2 × 2 mini-cubes which sharing a whole cube face, you can twist it 90 degrees in clockwise or counterclockwise direction, this twist operation is called one twist step.
Considering all faces of mini-cubes, there will be totally 24 faces painted in 6 different colors (Indexed from 0), and there will be exactly 4 faces painted in each kind of color. If 4 mini-cubes‘ faces of same color rely on same large cube face, we can call the large cube face as a completed face.
Now giving you an color arrangement of all 24 faces from a scrambled Pocket Cube, please tell us the maximum possible number of completed faces in no more than N twist steps.
Index of each face is shown as below:
Input
There will be several test cases. In each test case, there will be 2 lines. One integer N (1 ≤ N ≤ 7) in the first line, then 24 integers Ci separated by a single space in the second line. For index 0 ≤ i < 24, Ci is color of the corresponding face. We guarantee that the color arrangement is a valid state which can be achieved by doing a finite number of twist steps from an initial cube whose all 6 large cube faces are completed faces.
Output
For each test case, please output the maximum number of completed faces during no more than N twist step(s).
Sample Input
1
0 0 0 0 1 1 2 2 3 3 1 1 2 2 3 3 4 4 4 4 5 5 5 5
1
0 4 0 4 1 1 2 5 3 3 1 1 2 5 3 3 4 0 4 0 5 2 5 2
Sample Output
6
2
转魔方，bfs，一直找不到wa在哪里。这题思考魔方的旋转有点烦。
有6种旋转方式，而不是12种。。
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<vector>
5 #include<queue>
6 #include<algorithm>
7 #include<map>
8 #define INF 0x3f3f3f3f
9 #define M(a,b) memset(a,b,sizeof(a))
10
11 using namespace std;
12
13 int state[24];
14 int change[6][24]={
15 {6,1,12,3,5,11,16,7,8,9,4,10,18,13,14,15,20,17,22,19,0,21,2,23},
16 {20,1,22,3,10,4,0,7,8,9,11,5,2,13,14,15,6,17,12,19,16,21,18,23},
17 {1,3,0,2,23,22,4,5,6,7,10,11,12,13,14,15,16,17,18,19,20,21,9,8},
18 {2,0,3,1,6,7,8,9,23,22,10,11,12,13,14,15,16,17,18,19,20,21,5,4},
19 {0,1,11,5,4,16,12,6,2,9,10,17,13,7,3,15,14,8,18,19,20,21,22,23},
20 {0,1,8,14,4,3,7,13,17,9,10,2,6,12,16,15,5,11,18,19,20,21,22,23},
21 };
22
23 int N;
24 int ans;
25
26 struct tube
27 {
28 int d[24];
29 int cnt;
30 tube(){}
31 };
32 tube st;
33 tube trans(tube tmp,int k)
34 {
35 tube res;
36
37 for (int i=0; i<24; i++)
38 res.d[i]=tmp.d[change[k][i]];
39
40 return res;
41 }
42 int count(tube tmp)
43 {
44 int res=0;
45 if (tmp.d[0]==tmp.d[1] && tmp.d[1]==tmp.d[2] && tmp.d[2]==tmp.d[3]) res++;
46 if (tmp.d[8]==tmp.d[9] && tmp.d[14]==tmp.d[15] && tmp.d[8]==tmp.d[14]) res++;
47 if (tmp.d[6]==tmp.d[7] && tmp.d[7]==tmp.d[12] && tmp.d[12]==tmp.d[13]) res++;
48 if (tmp.d[4]==tmp.d[5] && tmp.d[5]==tmp.d[11] && tmp.d[11]==tmp.d[10]) res++;
49 if (tmp.d[16]==tmp.d[17] && tmp.d[17]==tmp.d[19] && tmp.d[19]==tmp.d[18]) res++;
50 if (tmp.d[20]==tmp.d[21] && tmp.d[21]==tmp.d[22] && tmp.d[22]==tmp.d[23]) res++;
51 return res;
52 }
53
54 void bfs()
55 {
56 queue<tube> q;
57 q.push(st);
58 // map<tube,int> mp;
59 // mp[st]=1;
60 ans=count(st);
61 while(!q.empty())
62 {
63 tube now=q.front();
64 q.pop();
65 tube tmp;
66 for (int i=0; i<6; i++)
67 {
68 tmp=trans(now,i);
69 tmp.cnt=now.cnt+1;
70 // if (mp.find(tmp)!=mp.end())
71 // {
72 // mp[tmp]=1;
73 // ans=max(ans,count(tmp));
74 // if (tmp.cnt<num) q.push(tmp);
75 // }
76 ans=max(ans,count(tmp));
77 if (tmp.cnt<N) q.push(tmp);
78 }
79 }
80 }
81
82 int main()
83 {
84 //init();
85 while(scanf("%d",&N)==1)
86 {
87 for(int i = 0;i<24;i++)
88 {
89 scanf("%d",&st.d[i]);
90 }
91 st.cnt=0;
92 bfs();
93 printf("%d\n",ans);
94 }
95 return 0;
96 }