At the beginning, the two dices may face different(which means there exist some i, ai ≠ bi). Ddy wants to make the two dices look the same from all directions(which means for all i, ai = bi) only by the following four rotation operations.(Please read the picture for more information)
For each case, the first line consists of six integers a1,a2,a3,a4,a5,a6, representing the numbers on dice A.
The second line consists of six integers b1,b2,b3,b4,b5,b6, representing the numbers on dice B.
题意:两个骰子,要把第一个骰子转到和第二个一样,有4种转法,求最少的次数
直接bfs,不过我用G++提交超时,用C++提交可以过,可能是使用了queue的原因。。。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<stdlib.h> 6 #include<algorithm> 7 #include<queue> 8 #include<map> 9 using namespace std; 10 struct Node 11 { 12 int a[7]; 13 int t; 14 }st,ed; 15 int vis[10][10][10][10][10][10]; 16 void bfs() 17 { 18 19 queue<Node>q; 20 q.push(st); 21 22 vis[st.a[0]][st.a[1]][st.a[2]][st.a[3]][st.a[4]][st.a[5]]=1; 23 Node t1,t2; 24 Node tmp; 25 while(!q.empty()) 26 { 27 t1=q.front(); 28 q.pop(); 29 if(t1.a[0]==ed.a[0] && t1.a[1]==ed.a[1]&& t1.a[2]==ed.a[2]&& t1.a[3]==ed.a[3]&& t1.a[4]==ed.a[4] && t1.a[5]==ed.a[5]) 30 { 31 printf("%d\n",t1.t); 32 return; 33 } 34 35 t2=t1; 36 t2.a[0]=t1.a[3]; 37 t2.a[1]=t1.a[2]; 38 t2.a[2]=t1.a[0]; 39 t2.a[3]=t1.a[1]; 40 if(vis[t2.a[0]][t2.a[1]][t2.a[2]][t2.a[3]][t2.a[4]][t2.a[5]]==0) 41 { 42 vis[t2.a[0]][t2.a[1]][t2.a[2]][t2.a[3]][t2.a[4]][t2.a[5]]=1; 43 t2.t=t1.t+1; 44 q.push(t2); 45 } 46 47 48 t2=t1; 49 t2.a[0]=t1.a[2]; 50 t2.a[1]=t1.a[3]; 51 t2.a[2]=t1.a[1]; 52 t2.a[3]=t1.a[0]; 53 if(vis[t2.a[0]][t2.a[1]][t2.a[2]][t2.a[3]][t2.a[4]][t2.a[5]]==0) 54 { 55 vis[t2.a[0]][t2.a[1]][t2.a[2]][t2.a[3]][t2.a[4]][t2.a[5]]=1; 56 t2.t=t1.t+1; 57 q.push(t2); 58 } 59 60 t2=t1; 61 t2.a[0]=t1.a[5]; 62 t2.a[1]=t1.a[4]; 63 t2.a[4]=t1.a[0]; 64 t2.a[5]=t1.a[1]; 65 if(vis[t2.a[0]][t2.a[1]][t2.a[2]][t2.a[3]][t2.a[4]][t2.a[5]]==0) 66 { 67 vis[t2.a[0]][t2.a[1]][t2.a[2]][t2.a[3]][t2.a[4]][t2.a[5]]=1; 68 t2.t=t1.t+1; 69 q.push(t2); 70 } 71 72 t2=t1; 73 t2.a[0]=t1.a[4]; 74 t2.a[1]=t1.a[5]; 75 t2.a[4]=t1.a[1]; 76 t2.a[5]=t1.a[0]; 77 if(vis[t2.a[0]][t2.a[1]][t2.a[2]][t2.a[3]][t2.a[4]][t2.a[5]]==0) 78 { 79 vis[t2.a[0]][t2.a[1]][t2.a[2]][t2.a[3]][t2.a[4]][t2.a[5]]=1; 80 t2.t=t1.t+1; 81 q.push(t2); 82 } 83 } 84 printf("-1\n"); 85 } 86 int main() 87 { 88 while(scanf("%d%d%d%d%d%d",&st.a[0],&st.a[1],&st.a[2],&st.a[3],&st.a[4],&st.a[5])==6) 89 { 90 scanf("%d%d%d%d%d%d",&ed.a[0],&ed.a[1],&ed.a[2],&ed.a[3],&ed.a[4],&ed.a[5]); 91 st.t=0; 92 memset(vis,0,sizeof(vis)); 93 bfs(); 94 } 95 return 0; 96 }