ZOJ 2477 Magic Cube 三阶魔方还原（IDA*）

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;}