sdutoj 2606 Rubik’s cube

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2606

 

Rubik’s cube

 

Time Limit: 2000ms   Memory limit: 65536K  有疑问？点这里^_^

题目描述

    Flabby is addicted to Rubik’s cube. He has many kinds of Rubik’s cube and plays them well. He can reorder a Rubik’s cube in a few seconds. He is famous for playing Rubik’s cube and attracts many young girls. One of his friends, Ant, is jealous of him and decides to let him be embarrassed in public. Ant makes a small Rubik’s cube by himself. This small magic is 2*2*2 and is drawn in blue and red, as shown in Figure 1. 
                                                              
    This Rubik’s cube can be rotated as well as other kinds of magic cube. Each of six faces can rotated clockwise and counterclockwise. Each 90 degree rotation on any face is counted as one step. The Rubik’s cube which has been reordered has only one color on each face.
     Ant is a clever boy. Sometimes, he can make a Rubik’s cube which can be reordered in a few steps. He can also make a Rubik’s cube which can’t be reordered in any way.
                                                            
                                                                 
Flabby knows what Ant thinks in his mind. He knows that you are a good programmer and asks you for help. Tell him whether this special Rubik’s cube can be reordered in a few steps.

输入

    In the input file, the first line is an integer T which is the number of test case. For each test case, there is 6 lines of integers describe the Rubik’s cube. For each line, there are four integers. Pij which is shown in Figure 3 corresponds to the jth integer of the ith line. If Pij is 0, it means the corresponding place is red. If Pij is 1, it means the corresponding place is blue. 

输出

    If the magic cube can be reordered in n steps, output the minimum n in a line. If the magic cube cannot be reordered, output “IMPOSSIBLE!” in a line. 

示例输入

3

0 0 0 0

0 1 0 1

1 1 0 0

1 0 1 0

1 1 0 0

1 1 1 1

0 0 1 1

1 1 0 0

0 1 0 1

1 0 0 0

1 1 1 1

0 1 0 0

1 0 1 1

0 1 0 0

0 0 0 0

1 1 1 1

1 0 0 0

0 1 1 1

示例输出

1

IMPOSSIBLE!

8

提示

 

来源

 2013年山东省第四届ACM大学生程序设计竞赛

示例程序

 

 

官方代码：

 

  1 /**

  2    23

  3    10

  4    23

  5    10

  6  010101

  7  323232

  8    01

  9    32

 10     6

 11     3

 12    412

 13     5

 14  */

 15 #include <iostream>

 16 #include <cstring>

 17 #include <cstdio>

 18 #include <cstdlib>

 19 using namespace std;

 20 char md[6][4];

 21 #define MAX 20000000

 22 bool flag[16][16][16][16][16][16];

 23 struct M

 24 {

 25     short int num[6];

 26     short int lv;

 27     void print()

 28     {

 29         int i;

 30         for(i = 0; i < 6; i++)

 31             printf("%d ",num[i]);

 32         printf("\n");

 33         return;

 34     }

 35 }queue[MAX];

 36 M (*funp[6])(M m);

 37 bool check(M m)

 38 {

 39     int i;

 40     for(i = 0; i < 6; i++)

 41     {

 42         if(m.num[i] != 15 && m.num[i] != 0)

 43             return false;

 44     }

 45     return true;

 46 }

 47 M turnY_L(M tm)

 48 {

 49     M m;

 50     m.num[0] = (tm.num[0] >> 1) | ((tm.num[0] & 1) << 3);

 51     m.num[1] = (tm.num[1] & 6) | ((tm.num[4] & 1) << 3) | ((tm.num[4] & 2) >> 1);

 52     m.num[2] = (tm.num[2] & 12) | ((tm.num[1] & 1) << 1) | ((tm.num[1] & 8) >> 3);

 53     m.num[3] = (tm.num[3] & 9) | ((tm.num[2] & 2) << 1) | ((tm.num[2] & 1) << 1);

 54     m.num[4] = (tm.num[4] & 12) | ((tm.num[3] & 4) >> 1) | ((tm.num[3] & 2) >> 1);

 55     m.num[5] = tm.num[5];

 56     return m;

 57 }

 58 M turnY_R(M tm)

 59 {

 60     M m;

 61     m.num[0] = ((tm.num[0] & 7) << 1) | ((tm.num[0] & 8) >> 3);

 62     m.num[1] = (tm.num[1] & 6) | ((tm.num[2] & 2) >> 1) | ((tm.num[2] & 1) << 3);

 63     m.num[2] = (tm.num[2] & 12) | ((tm.num[3] & 4) >> 1) | ((tm.num[3] & 2) >> 1);

 64     m.num[3] = (tm.num[3] & 9) | ((tm.num[4] & 1) << 1) | ((tm.num[4] & 2) << 1);

 65     m.num[4] = (tm.num[4] & 12) | ((tm.num[1] & 1) << 1) | ((tm.num[1] & 8) >> 3);

 66     m.num[5] = tm.num[5];

 67     return m;

 68 }

 69 

 70 /*------------------------------------------------------------------------------------*/

 71 

 72 M turnX_L(M tm)

 73 {

 74     M m;

 75     m.num[1] = (tm.num[1] >> 1) | ((tm.num[1] & 1) << 3);

 76     m.num[0] = (tm.num[0] & 9) | ((tm.num[2] & 1) << 2) | ((tm.num[2] & 8) >> 2);

 77     m.num[2] = (tm.num[2] & 6) | ((tm.num[5] & 9));

 78     m.num[3] = tm.num[3];

 79     m.num[4] = (tm.num[4] & 9) | ((tm.num[0] & 6));

 80     m.num[5] = (tm.num[5] & 6) | ((tm.num[4] & 4) >> 2) | ((tm.num[4] & 2) << 2);

 81     return m;

 82 }

 83 M turnX_R(M tm)

 84 {

 85     M m;

 86     m.num[1] = ((tm.num[1] & 7) << 1) | ((tm.num[1] & 8) >> 3);

 87     m.num[0] = (tm.num[0] & 9) | ((tm.num[4] & 6));

 88     m.num[2] = (tm.num[2] & 6) | ((tm.num[0] & 4) >> 2) | ((tm.num[0] & 2) << 2);

 89     m.num[3] = tm.num[3];

 90     m.num[4] = (tm.num[4] & 9) | ((tm.num[5] & 8) >> 2) | ((tm.num[5] & 1) << 2);

 91     m.num[5] = (tm.num[5] & 6) | ((tm.num[2] & 9));

 92     return m;

 93 }

 94 

 95 M turnZ_L(M tm)

 96 {

 97     M m;

 98     m.num[4] = (tm.num[4] >> 1) | ((tm.num[4] & 1) << 3);

 99     m.num[0] = (tm.num[0] & 3) | (tm.num[1] & 12);

100     m.num[1] = (tm.num[1] & 3) | (tm.num[5] & 12);

101     m.num[2] = tm.num[2];

102     m.num[3] = (tm.num[3] & 3) | (tm.num[0] & 12);

103     m.num[5] = (tm.num[5] & 3) | (tm.num[3] & 12);

104     return m;

105 }

106 M turnZ_R(M tm)

107 {

108     M m;

109     m.num[4] = ((tm.num[4] & 7)<< 1) | ((tm.num[4] & 8) >> 3);

110     m.num[0] = (tm.num[0] & 3) | (tm.num[3] & 12);

111     m.num[1] = (tm.num[1] & 3) | (tm.num[0] & 12);

112     m.num[2] = tm.num[2];

113     m.num[3] = (tm.num[3] & 3) | (tm.num[5] & 12);

114     m.num[5] = (tm.num[5] & 3) | (tm.num[1] & 12);

115     return m;

116 }

117 

118 

119 void record_flag(int num1,int num2,int num3,int num4,int num5,int num6)

120 {

121     //printf("%d %d %d %d %d %d\n",num1,num2,num3,num4,num5,num6);

122     flag[num1][num2][num3][num4][num5][num6] = 1;

123     flag[num4][num1][(num3>>1)|((num3&1)<<3)][num6][((num5&7)<<1)+((num3&8)>>3)][num2] = 1;

124     flag[num6][num4][((num3&3)<<2)+((num3&12)>>2)][num2][((num5&3)<<2)+((num5&12)>>2)][num1] = 1;

125     flag[num2][num6][((num3&7)<<1)|((num3&8)>>3)][num1][(num5>>1)+((num5&1)<<3)][num4] = 1;

126     return;

127 }

128 int Search(M m)

129 {

130     funp[0] = turnX_L;

131     funp[1] = turnX_R;

132     funp[2] = turnY_L;

133     funp[3] = turnY_R;

134     funp[4] = turnZ_L;

135     funp[5] = turnZ_R;

136     M tmp,tm;

137     int front,rear,i;

138     front = rear = 0;

139     memset(flag,0,sizeof(flag));

140     m.lv = 0;

141     queue[rear++] = m;

142     record_flag(m.num[0],m.num[1],m.num[2],m.num[3],m.num[4],m.num[5]);

143     while(front < rear)

144     {

145         tmp = queue[front++];

146         if(check(tmp))

147             return tmp.lv;

148         for(i = 0; i < 6; i++)

149         {

150             tm = funp[i](tmp);

151             tm.lv = tmp.lv + 1;

152             if(flag[tm.num[0]][tm.num[1]][tm.num[2]][tm.num[3]][tm.num[4]][tm.num[5]] == 0)

153             {

154                 queue[rear++] = tm;

155                 record_flag(tm.num[0],tm.num[1],tm.num[2],tm.num[3],tm.num[4],tm.num[5]);

156             }

157         }

158     }

159     return -1;

160 }

161 

162 int main()

163 {

164     int T,i,n1,n2,n3,n4,ans;

165     freopen("data.in","r",stdin);

166     freopen("data.out","w",stdout);

167     scanf("%d",&T);

168     M m;

169     while(T--)

170     {

171         for(i = 0; i < 6; i++)

172         {

173             scanf("%d%d%d%d",&n1,&n2,&n3,&n4);

174             m.num[i] = n1 + (n2 << 1) + (n3 << 3) + (n4 << 2);

175           //  printf("%d\n",m.num[i]);

176         }

177         ans = Search(m);

178         if(ans != -1)

179             printf("%d\n", ans);

180         else

181             printf("IMPOSSIBLE!\n");

182     }

183     return 0;

184 }

View Code

 

 

 

官方数据生成代码：

 

 1 #include <iostream>

 2 #include <cstring>

 3 #include <cstdio>

 4 #include <ctime>

 5 #include <cstdlib>

 6 using namespace std;

 7 int a[24] = {0,0,1,1,1,1,0,0,0,1,0,1,1,0,1,0,1,1,0,1,0,1,0,0};

 8 int main()

 9 {

10     int t1,t2,i;

11     freopen("data.in","w",stdout);

12     int T = 10;

13     printf("%d\n",T);

14     srand(0);

15     while(T--)

16     {

17 

18         t1 = rand()%24;

19         do

20         {

21             t2 = ((rand()%24)*(rand()%24))%24;

22         }while(a[t1]==a[t2]);

23         a[t1] = a[t1] ^ a[t2];

24         a[t2] = a[t1] ^ a[t2];

25         a[t1] = a[t1] ^ a[t2];

26         for(i = 0; i < 24; i++)

27         {

28             printf("%d",a[i]);

29             if(i % 4 == 3)

30                 printf("\n");

31             else

32                 printf(" ");

33         }

34         printf("\n");

35     }

36     return 0;

37 }

View Code