题目大意是提供四个六面颜色各异的立方体，让你垒起来，得到的长方体除掉top和bottom的四面每一面小格子颜色都不重复。
思路就是找出每个立方体能够被用作大长方体四面的部分，然后慢慢用穷举法拼，当然其间有剪枝。

开始犯了个算是致命的错误————在找每个立方体能被用作大长方体四面部分的函数，也就是get_round函数，起初只考虑了一个方向,结果输出空白。这个陷阱挺坏的……41-58行加上后，输出应该是正确的了。

[c highlight=”41-58,”]
#include <stdio.h>

#define R 1
#define B 2
#define G 4
#define Y 8

#define FRONT 0
#define BACK 1
#define LEFT 2
#define RIGHT 3
#define TOP 4
#define BOTTOM 5

static int CUBEA[6] = {R,B,G,Y,B,Y};
static int CUBEB[6] = {R,G,G,Y,B,B};
static int CUBEC[6] = {Y,B,R,G,Y,R};
static int CUBED[6] = {Y,G,B,R,R,R};

void get_round(int * cube, int* round, int side){
switch(side){
case 0:
round[0] = cube[FRONT];
round[1] = cube[RIGHT];
round[2] = cube[BACK];
round[3] = cube[LEFT];
return;
case 1:
round[0] = cube[FRONT];
round[1] = cube[TOP];
round[2] = cube[BACK];
round[3] = cube[BOTTOM];
return;
case 2:
round[0] = cube[TOP];
round[1] = cube[RIGHT];
round[2] = cube[BOTTOM];
round[3] = cube[LEFT];
return;

case 3:
round[3] = cube[FRONT];
round[2] = cube[RIGHT];
round[1] = cube[BACK];
round[0] = cube[LEFT];
return;
case 4:
round[3] = cube[FRONT];
round[2] = cube[TOP];
round[1] = cube[BACK];
round[0] = cube[BOTTOM];
return;;
case 5:
round[3] = cube[TOP];
round[2] = cube[RIGHT];
round[1] = cube[BOTTOM];
round[0] = cube[LEFT];
return;
}

}
char* to_color(int i){
switch(i){
case 1: return “R”;
case 2: return “B”;
case 4: return “G”;
case 8: return “Y”;
default: printf(“errorinput:%d\n”,i); return “ERR”;
}
}
void print_result(int* cube, int side, int off){
int r[4]={0};
get_round(cube,r,side);
printf(“%s,%s,%s,%s”, to_color(r[(0 + off) % 4]), to_color(r[(1 + off) % 4]), to_color(r[(2 + off) % 4]), to_color(r[(3 + off) % 4]));
}

void check_rounds(int as,int* a,int bs, int* b, int cs, int* c, int ds,int*d)
{
int has = 1;
for(int ai=0;ai<4;ai++){
for(int bi=0;bi<4;bi++){
if(a[ai]!=b[bi])
for(int ci=0;ci<4;ci++){
if(a[ai]!=c[ci] && b[bi] != c[ci])
for(int di=0;di<4;di++){
has = 1;
for(int i=0;i<4;i++){
if((a[(ai + i) % 4]|b[(bi + i) % 4]|c[(ci + i) % 4]|d[(di + i) % 4])!=15) {
has = 0; break;
}
}

if(has == 1){
printf(“\t1-“);
print_result(CUBEA,as,ai);
printf(“\n\t2-“);
print_result(CUBEB,bs,bi);
printf(“\n\t3-“);
print_result(CUBEC,cs,ci);
printf(“\n\t4-“);
print_result(CUBED,ds,di);
printf(“\n\t———\n”);
return;
}
}
}
}
}
}

int main(int argc, char** argv){
int rounda[4] = {0};
int roundb[4] = {0};
int roundc[4] = {0};
int roundd[4] = {0};

for(int i=0;i<6;i++){
get_round(CUBEA,rounda,i);
for(int ii = 0;ii<6;ii++){
get_round(CUBEB,roundb,ii);
for(int iii=0;iii<6;iii++){
get_round(CUBEC,roundc,iii);
for(int iiii=0;iiii<6;iiii++){
get_round(CUBED, roundd, iiii);
check_rounds(i, rounda, ii, roundb, iii, roundc, iiii, roundd);
}
}
}
}
return 0;

}
[/c]

结果如下：

1-R,B,B,Y
2-G,Y,R,G
3-B,G,Y,R
4-Y,R,G,B
———
1-Y,B,B,R
2-G,R,Y,G
3-R,Y,G,B
4-B,G,R,Y
———