UVALive 3401 彩色立方體

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1402

http://7xjob4.com1.z0.glb.clouddn.com/53f6b2526cc5a59ec7881a8fd6d899bd

題意：將n個原有顏色的立方體塗儘可能少次使立方體都相同；

思路：枚舉除第一個面外每一個立方體的姿態（24種），姿態由旋轉方式先處理獲得，再枚舉每個對應面記錄要塗的數量。

處理代碼：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 int lleft[7]={0,2,6,3,4,1,5};
 5 int up[7]={0,4,2,1,6,5,3};
 6 
 7 void rot(int *T,int *p)
 8 {
 9     int q[7];
10     for(int i=1;i<=6;i++)
11     {
12         q[i]=p[i];
13     }
14 
15     for(int i=1;i<=6;i++)
16     {
17         p[i]=q[T[i]];
18     }
19 }
20 
21 int main()
22 {
23     int p0[7]={0,1,2,3,4,5,6};
24 
25     printf("int dice[25][7]= {\n{0},\n");
26     for(int i=1;i<=6;i++)
27     {
28         int p[7];
29         for(int j=0;j<=6;j++)
30             p[j]=p0[j];
31 
32         if(i==1) rot(up,p);
33         if(i==2) {  rot(lleft,p); rot(up,p); }
34 
35         if(i==4) {  rot(up,p);rot(up,p); }
36         if(i==5) {  rot(lleft,p);rot(lleft,p);rot(lleft,p);rot(up,p); }
37         if(i==6) {  rot(lleft,p);rot(lleft,p);rot(up,p); }
38 
39         for(int j=1;j<=4;j++)
40         {
41             printf("{%d,%d,%d,%d,%d,%d,%d},",p[0],p[1],p[2],p[3],p[4],p[5],p[6]);
42             rot(lleft,p);
43         }
44         printf("\n");
45     }
46     printf("}; \n");
47 }

View Code

計算代碼：

  1 #include <bits/stdc++.h>
  2 #include <iostream>
  3 using namespace std;
  4 
  5 int dice[25][7]= {
  6 {0},
  7 {0,4,2,1,6,5,3},{0,2,3,1,6,4,5},{0,3,5,1,6,2,4},{0,5,4,1,6,3,2},
  8 {0,4,6,2,5,1,3},{0,6,3,2,5,4,1},{0,3,1,2,5,6,4},{0,1,4,2,5,3,6},
  9 {0,1,2,3,4,5,6},{0,2,6,3,4,1,5},{0,6,5,3,4,2,1},{0,5,1,3,4,6,2},
 10 {0,6,2,4,3,5,1},{0,2,1,4,3,6,5},{0,1,5,4,3,2,6},{0,5,6,4,3,1,2},
 11 {0,4,1,5,2,6,3},{0,1,3,5,2,4,6},{0,3,6,5,2,1,4},{0,6,4,5,2,3,1},
 12 {0,4,5,6,1,2,3},{0,5,3,6,1,4,2},{0,3,2,6,1,5,4},{0,2,4,6,1,3,5},
 13 };
 14 
 15 int n,ans;
 16 int color[5][7],state[5];
 17 vector <string> colorname;
 18 
 19 int colorid(char str[])
 20 {
 21     int i,j;
 22     string s(str);
 23     int m=colorname.size();
 24     for(i=0;i<m;i++)
 25     {
 26         if(colorname[i]==s)
 27         {
 28             return i;
 29         }
 30     }
 31     colorname.push_back(s);
 32     return m;
 33 }
 34 
 35 void cal()
 36 {
 37     int i,j;
 38     int num=0,maxnum;
 39     map <int,int> cn;
 40     for(j=1;j<=6;j++)
 41     {
 42         cn.clear();
 43         maxnum=0;
 44         for(i=1;i<=n;i++)
 45         {
 46             int cnam=color[i][dice[state[i]][j]];
 47             cn[cnam]++;
 48             if(cn[cnam]>maxnum)
 49                 maxnum=cn[cnam];
 50         }
 51         num+=(n-maxnum);
 52     }
 53     if(ans>num)
 54         ans=num;
 55 }
 56 
 57 void dfs(int m)
 58 {
 59     int i,j;
 60     if(m==n)
 61     {
 62         cal();
 63         return;
 64     }
 65     for(i=1;i<=24;i++)
 66     {
 67         state[m+1]=i;
 68         dfs(m+1);
 69     }
 70 }
 71 int main()
 72 {
 73     char str[30];
 74     int i,j;
 75     while(scanf("%d",&n)!=EOF && n!=0)
 76     {
 77         colorname.clear();
 78         for(i=1;i<=n;i++)
 79         {
 80             for(j=1;j<=6;j++)
 81             {
 82                 scanf("%s",str);
 83                 color[i][j]=colorid(str);
 84             }
 85         }
 86 
 87         if(n==1)
 88         {
 89             ans=0;
 90         }
 91         else
 92         {
 93             ans=24;
 94             state[1]=9;
 95             dfs(1);
 96         }
 97 
 98         printf("%d\n",ans);
 99     }
100     return 0;
101 }
102 
103 /*
104 for(i=1;i<=n;i++)
105         {
106             for(j=1;j<=6;j++)
107             {
108                 printf("%s ",color[i][j]);
109             }
110             printf("\n");
111         }
112         if(strcmp(color[1][2],color[2][3])==0)
113         {
114             printf("yes\n");
115         }
116 */

View Code