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2013区域赛长沙赛区现场赛 K - Pocket Cube

时间:2014-10-06 02:03:49      阅读:376      评论:0      收藏:0      [点我收藏+]

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K - Pocket Cube
Time Limit:10000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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. 

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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: 

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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 }

 

2013区域赛长沙赛区现场赛 K - Pocket Cube

标签:des   style   blog   http   color   io   os   ar   for   

原文地址:http://www.cnblogs.com/haohaooo/p/4007845.html

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