LA 3401 - Colored Cubes

color₁ color₂ color₃ color₄ color₅ color₆

corresponds to a cube colored as shown in Figure 6.

The end of the input is indicated by a line containing a single zero. It is not a dataset nor a part of a dataset.

Figure 2: Identically colored cubes

Figure 3: cubes that are not identically colored

Figure 4: An example of recoloring

Figure 5: Numbering of faces Figure 6: Coloring

Output 

For each dataset, output a line containing the minimum number of faces that need to be repainted to make the set of cub es identically colored.

Sample Input 

3 scarlet green blue yellow magenta cyan blue pink green magenta cyan lemon purple red blue yellow cyan green 2 red green blue yellow magenta cyan cyan green blue yellow magenta red 2 red green gray gray magenta cyan cyan green gray gray magenta red 2 red green blue yellow magenta cyan magenta red blue yellow cyan green 3 red green blue yellow magenta cyan cyan green blue yellow magenta red magenta red blue yellow cyan green 3 blue green green green green blue green blue blue green green green green green green green green sea-green 3 red yellow red yellow red yellow red red yellow yellow red yellow red red red red red red 4 violet violet salmon salmon salmon salmon violet salmon salmon salmon salmon violet violet violet salmon salmon violet violet violet violet violet violet salmon salmon 1 red green blue yellow magenta cyan 4 magenta pink red scarlet vermilion wine-red aquamarine blue cyan indigo sky-blue turquoise-blue blond cream chrome-yellow lemon olive yellow chrome-green emerald-green green olive vilidian sky-blue 0

Sample Output 

4

2

0

0

2

3

4

4

0

16

#include <iostream>#include <stdio.h>#include <map>#include <queue>#include <cstring>#include <string.h>using namespace std;#define FOR(i,n) for(int i=1; i<=n; i++)int all[24][6] = {{1,2,3,4,5,6},{2,3,4,1,5,6},{3,4,1,2,5,6},{4,1,2,3,5,6},{5,1,6,3,2,4},{1,6,3,5,2,4},{6,3,5,1,2,4},{3,5,1,6,2,4},{2,5,4,6,1,3},{5,4,6,2,1,3},{4,6,2,5,1,3},{6,2,5,4,1,3},{1,4,3,2,6,5},{4,3,2,1,6,5},{3,2,1,4,6,5},{2,1,4,3,6,5},{6,1,5,3,4,2},{1,5,3,6,4,2},{5,3,6,1,4,2},{3,6,1,5,4,2},{4,5,2,6,3,1},{5,2,6,4,3,1},{2,6,4,5,3,1},{6,4,5,2,3,1},};map<string,int>mID;int n, id, ans;int MoFang[5][7];int sum[30];int get_temp(){    int rt = 0;    FOR(j,6)    {        int s = 0;        memset(sum,0,sizeof sum);        FOR(i,n)        {            sum[MoFang[i][j]]++;            s = max(s,sum[MoFang[i][j]]) ;        }        rt += n - s;    }    return rt;}void dfs(int xb){    if(xb==n+1)    {        int temp = get_temp();                if(temp<ans)         {            ans = temp;        }        return;    }    for(int i=0; i<24; i++)    {        int bc[7];        FOR(j,6) bc[j] = MoFang[xb][j];        FOR(j,6) MoFang[xb][j] = bc[all[i][j-1]];        dfs(xb+1);        FOR(j,6) MoFang[xb][j] = bc[j];    }}int shunxu[]={0,3,4,5,6,2,1};int main(){    while(scanf("%d",&n)!=EOF)    {        if(n==0)break;        char s0[10];        int cl;        id = 1;        mID.clear();        FOR(i,n) FOR(j,6)         {            scanf("%s",s0);            string s="";            int len = strlen(s0);            for(int k=0; k<len ; k++)            {                s += s0[k];            }            s[len] ='\0';                  if(mID[s])            {                cl = mID[s];            }else{                cl = id++;                mID[s] = cl;            }            MoFang[i][shunxu[j]] = cl;        }                ans = 6*n;        dfs(2);        printf("%d\n",ans);    }    return 0;}/*map<string,bool>QuChong;int top;void dfs(char g[]){    string s="";    char c;    int i;    for(i=1; i<=6; i++)    {        s += g[i];    }    s[6] ='\0';    if(QuChong[s])    {        return;    }else{               QuChong[s] = true;        printf("{");        for(i=1; i<6; i++)        printf("%c,",g[i]);        printf("%c},",g[6]);        top++;    }    char next[10];    memcpy(next, g, 8);     c = next[1];         for(i=1; i<4; i++)next[i] = next[i+1];         next[4] = c;    dfs(next);        memcpy(next, g, 8);    c = next[1];    next[1] = next[5]; next[5] = next[3]; next[3] = next[6]; next[6] = c;    dfs(next);}void get_all(){    QuChong.clear();    top =0;    printf("int all[24][6] = {");    char g[10]={'0','1','2','3','4','5','6','\0'};    dfs(g);    printf("}\n");    cout<<endl<<(sizeof(g))<<endl;    cout<<endl<<top<<endl;}*/