转载请注明出处，谢谢 http://blog.csdn.net/ACM_cxlove?viewmode=contents by—cxlove
三阶魔方还原。因为只搜5层，所以使用IDA*搜索。由于每次旋转，每面中心颜色总不变，也就确定了最终的状态，找出每个面中与中间颜色不同的个数的最大值，其中每次旋转会更改每个面的3个位置的颜色，所以 就是（最大值+2）/3。
总共有12种旋转，找到其中的对应方式，使用转动数组就非常方便了，不过数组的初始化是相当蛋疼的工作，最好手头有个三阶魔方。顺时针与逆时针刚好对应。
#include<iostream>#include<cstdio>#include<cstring>using namespace std;//表示每个面除中间的另外8个的位置short int cen[6][8]={ {0,1,2,3,5,6,7,8},{9,10,11,21,23,33,34,35},{12,13,14,24,26,36,37,38}, {15,16,17,27,29,39,40,41},{18,19,20,30,32,42,43,44},{45,46,47,48,50,51,52,53}};//转换数组，12种变换，两两对应，每次转换会更改20个位置short int change[12][20]={ {11,23,35,34,33,21,9,10,51,48,45,36,24,12,6,3,0,20,32,44}, {9,10,11,23,35,34,33,21,36,24,12,6,3,0,20,32,44,51,48,45}, {14,13,26,38,37,36,24,12,45,46,47,39,27,15,8,7,6,11,23,35}, {12,24,13,14,26,38,37,36,39,27,15,8,7,6,11,23,35,45,46,47}, {17,29,41,40,39,27,15,16,47,50,53,42,30,18,2,5,8,14,26,38}, {15,16,17,29,41,40,39,27,42,30,18,2,5,8,14,26,38,47,50,53}, {18,19,20,32,44,43,42,30,53,52,51,33,21,9,0,1,2,17,29,41}, {42,30,18,19,20,32,44,43,33,21,9,0,1,2,17,29,41,53,52,51}, {0,1,2,5,8,7,6,3,12,13,14,15,16,17,18,19,20,9,10,11}, {6,3,0,1,2,5,8,7,15,16,17,18,19,20,9,10,11,12,13,14}, {45,46,47,50,53,52,51,48,44,43,42,41,40,39,38,37,36,35,34,33}, {51,48,45,46,47,50,53,52,41,40,39,38,37,36,35,34,33,44,43,42}};char a[54]; //初始状态int depth; //迭代加深搜索的层数bool flag; //是否有解int centre[6]={4,22,25,28,31,49};//每个面中心坐标int get_h(char *maze){ int ret=0; for(int i=0;i<6;i++){ int cnt=0; for(int j=0;j<8;j++) if(maze[cen[i][j]]!=maze[centre[i]]) cnt++; ret=max(ret,cnt); } return (ret+2)/3;}int ans[10];//调试用的，输出当前的形状void debug(char *maze){ int k=0; for(int i=0;i<3;i++){ printf(" "); for(int j=0;j<3;j++) printf("%c ",maze[k++]); printf("\n"); } for(int i=0;i<3;i++){ for(int j=0;j<12;j++) printf("%c ",maze[k++]); printf("\n"); } for(int i=0;i<3;i++){ printf(" "); for(int j=0;j<3;j++) printf("%c ",maze[k++]); printf("\n"); }}void IDAstar(int tmp_depth,char *b,int father){ if(flag) return; //A*剪枝 if(get_h(b)>tmp_depth) return; if(tmp_depth==0){ flag=true; return; } for(int i=0;i<12;i++){ if(flag) return; if((i^father)==1) continue; char tmp[54]; memcpy(tmp,b,54*sizeof(char)); ans[tmp_depth]=i; //转换 for(int j=0;j<20;j++) tmp[change[i][j]]=b[change[i^1][j]];; IDAstar(tmp_depth-1,tmp,i); }}char get_in(){ char ch; while(1){ ch=getchar(); if(ch>='a'&&ch<='z') return ch; }}int main(){ int t; scanf("%d",&t); while(t--){ for(int i=0;i<54;i++) a[i]=get_in(); flag=false; int Init=get_h(a); if(Init==0){ printf("0\n"); continue; } for(depth=Init;depth<=5;depth++){ IDAstar(depth,a,-1); if(flag){ printf("%d\n",depth); printf("%d %d",ans[depth]/2,(ans[depth]&1)?-1:1); for(int j=depth-1;j>0;j--) printf(" %d %d",ans[j]/2,(ans[j]&1)?-1:1); printf("\n"); break; } } if(!flag) printf("-1\n"); } return 0;}
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