Problem Description
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
Source
题目意思：
一个正方形，每个面有一个数字而且不相同（最多为6最小为1），正方形可以向左、右、前、后转，给出正方形两个状态，问最少转多少次可以从前面的状态转换成后面的状态，若不能输出-1。
思路：
正方形最多有6!种状态，由前面状态推到最后的状态，很明显bfs。
题目解析：因为是网络赛没看题就觉得是几何题，觉得做不了，之后看交的多了，就看了一下题，没想到是bfs模板水题。
#include#include #include #include #include using namespace std;int v[7][7][7][7][7][7];struct node{ int ff[7]; int ans;};struct node t,f;int a[7],b[7];int bfs(){ queue q; for(int i=1; i<=6; i++) t.ff[i]=a[i]; t.ans=0; q.push(t); v[t.ff[1]][t.ff[2]][t.ff[3]][t.ff[4]][t.ff[5]][t.ff[6]]=1; while(!q.empty()) { t=q.front(); q.pop(); if(t.ff[1]==b[1]&&t.ff[2]==b[2]&&t.ff[3]==b[3]&&t.ff[4]==b[4]&&t.ff[5]==b[5]&&t.ff[6]==b[6]) { printf("%d\n",t.ans); return 1; } for(int i=1; i<=4; i++) { if(i==1) { f.ff[1]=t.ff[4]; f.ff[2]=t.ff[3]; f.ff[3]=t.ff[1]; f.ff[4]=t.ff[2]; f.ff[5]=t.ff[5]; f.ff[6]=t.ff[6]; if(v[f.ff[1]][f.ff[2]][f.ff[3]][f.ff[4]][f.ff[5]][f.ff[6]]==0) { f.ans=t.ans+1; v[f.ff[1]][f.ff[2]][f.ff[3]][f.ff[4]][f.ff[5]][f.ff[6]]=1; q.push(f); } } else if(i==2) { f.ff[1]=t.ff[3]; f.ff[2]=t.ff[4]; f.ff[3]=t.ff[2]; f.ff[4]=t.ff[1]; f.ff[5]=t.ff[5]; f.ff[6]=t.ff[6]; if(v[f.ff[1]][f.ff[2]][f.ff[3]][f.ff[4]][f.ff[5]][f.ff[6]]==0) { f.ans=t.ans+1; v[f.ff[1]][f.ff[2]][f.ff[3]][f.ff[4]][f.ff[5]][f.ff[6]]=1; q.push(f); } } else if(i==3) { f.ff[1]=t.ff[6]; f.ff[2]=t.ff[5]; f.ff[3]=t.ff[3]; f.ff[4]=t.ff[4]; f.ff[5]=t.ff[1]; f.ff[6]=t.ff[2]; if(v[f.ff[1]][f.ff[2]][f.ff[3]][f.ff[4]][f.ff[5]][f.ff[6]]==0) { f.ans=t.ans+1; v[f.ff[1]][f.ff[2]][f.ff[3]][f.ff[4]][f.ff[5]][f.ff[6]]=1; q.push(f); } } else if(i==4) { f.ff[1]=t.ff[5]; f.ff[2]=t.ff[6]; f.ff[3]=t.ff[3]; f.ff[4]=t.ff[4]; f.ff[5]=t.ff[2]; f.ff[6]=t.ff[1]; if(v[f.ff[1]][f.ff[2]][f.ff[3]][f.ff[4]][f.ff[5]][f.ff[6]]==0) { f.ans=t.ans+1; v[f.ff[1]][f.ff[2]][f.ff[3]][f.ff[4]][f.ff[5]][f.ff[6]]=1; q.push(f); } } } } return 0;}int main(){ int F; while(scanf("%d%d%d%d%d%d",&a[1],&a[2],&a[3],&a[4],&a[5],&a[6])!=EOF) { for(int i=1; i<=6; i++) scanf("%d",&b[i]); memset(v,0,sizeof(v)); F=bfs(); if(F==0) printf("-1\n"); } return 0;}