ZOJ 2477 Magic Cube 三阶魔方还原（IDA*）

转载请注明出处，谢谢http://blog.youkuaiyun.com/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;
}