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There are 2 special dices on the table. On each face of the dice, a distinct number was written. Consider a 1.a 2,a 3,a 4,a 5,a 6 to be numbers written on top face, bottom face, left face, right face, front face and back face of dice A. Similarly, consider b 1.b 2,b 3,b 4,b 5,b 6 to be numbers on specific faces of dice B. It’s guaranteed that all numbers written on dices are integers no smaller than 1 and no more than 6 while a i ≠ a j and b i ≠ b j for all i ≠ j. Specially, sum of numbers on opposite faces may not be 7.

At the beginning, the two dices may face different(which means there exist some i, a i ≠ b i). Ddy wants to make the two dices look the same from all directions(which means for all i, a i = b i) only by the following four rotation operations.(Please read the picture for more information)

Now Ddy wants to calculate the minimal steps that he has to take to achieve his goal. Input There are multiple test cases. Please process till EOF.

For each case, the first line consists of six integers a 1,a 2,a 3,a 4,a 5,a 6, representing the numbers on dice A.

The second line consists of six integers b 1,b 2,b 3,b 4,b 5,b 6, representing the numbers on dice B.

Output For each test case, print a line with a number representing the answer. If there’s no way to make two dices exactly the same, output -1. Sample Input 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 5 6 4 3 1 2 3 4 5 6 1 4 2 5 3 6 Sample Output 0 3 -1 两个筛子，如果可以通过绕着底部四条边使他们成为一样的，就输出最少的步数。如果不能的话就输出-1。 输出-1的情况就是相对的两个面不是一样的数字，其余的情况在六步内一定可以输出最少的步数。 读入的时候，依次是上下左右前后。 直接广搜给上就行。 代码如下：

#include
   
    using namespace std;int vis1[10],vis2[10];map
    
     mp;int a[10],b[10];struct node{
   	int ans[10];	int a;};bool check(node c){
   	for(int i=1;i<=6;i++) if(c.ans[i]!=b[i]) return false;	return true;}int bfs(){
   	mp.clear();	queue
     
       s;	node x,z;	string cnt="";	for(int i=1;i<=6;i++) x.ans[i]=a[i],cnt+=a[i]+'0';	x.a=0;	mp[cnt]=1;	s.push(x);	while(s.size())	{
   		node y=s.front();		s.pop();		if(check(y)) return y.a;		z.ans[1]=y.ans[6];z.ans[2]=y.ans[5];z.ans[3]=y.ans[3];z.ans[4]=y.ans[4];z.ans[5]=y.ans[1];z.ans[6]=y.ans[2];		cnt="";for(int i=1;i<=6;i++) cnt+=z.ans[i]+'0';if(!mp[cnt]) z.a=y.a+1,mp[cnt]=1,s.push(z);		z.ans[1]=y.ans[5];z.ans[2]=y.ans[6];z.ans[3]=y.ans[3];z.ans[4]=y.ans[4];z.ans[5]=y.ans[2];z.ans[6]=y.ans[1];		cnt="";for(int i=1;i<=6;i++) cnt+=z.ans[i]+'0';if(!mp[cnt]) z.a=y.a+1,mp[cnt]=1,s.push(z);		z.ans[1]=y.ans[4];z.ans[2]=y.ans[3];z.ans[3]=y.ans[1];z.ans[4]=y.ans[2];z.ans[5]=y.ans[5];z.ans[6]=y.ans[6];		cnt="";for(int i=1;i<=6;i++) cnt+=z.ans[i]+'0';if(!mp[cnt]) z.a=y.a+1,mp[cnt]=1,s.push(z);		z.ans[1]=y.ans[3];z.ans[2]=y.ans[4];z.ans[3]=y.ans[2];z.ans[4]=y.ans[1];z.ans[5]=y.ans[5];z.ans[6]=y.ans[6];		cnt="";for(int i=1;i<=6;i++) cnt+=z.ans[i]+'0';if(!mp[cnt]) z.a=y.a+1,mp[cnt]=1,s.push(z);	}	return -1;}int main(){
   	while(scanf("%d",&a[1])!=EOF)	{
   		for(int i=2;i<=6;i++) scanf("%d",&a[i]);		for(int i=1;i<=6;i++) scanf("%d",&b[i]);		vis1[a[1]]=vis1[a[2]]=1;		vis1[a[3]]=vis1[a[4]]=2;		vis1[a[5]]=vis1[a[6]]=3;		vis2[b[1]]=vis2[b[2]]=1;		vis2[b[3]]=vis2[b[4]]=2;		vis2[b[5]]=vis2[b[6]]=3;		int flag=1;		for(int i=1;i<=6;i+=2)		{
   			if(vis2[a[i]]!=vis2[a[i+1]])			{
   				flag=0;				break;			}		}		if(flag==0) printf("-1\n");		else printf("%d\n",bfs());	}}
     
    
   

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