題干：

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

題目大意：

給出兩個正方體，每個正方體都有6種顏色，問是否可以經過左轉，右轉，前轉，后轉四種轉法讓兩個正方體看起來一樣(對應面顏色相同)，最少轉幾次。

解題報告：

?直接bfs。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<stack> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define FF first #define SS second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 2e5 + 5; vector<int> aa; set<vector<int> > ss; struct Node {vector<int> vv;int s; }; int bfs(vector<int> st) {queue<Node> q;q.push({st,0});while(q.size()) {Node cur = q.front();q.pop();if(cur.vv == aa) return cur.s;if(ss.count(cur.vv)) continue;ss.insert(cur.vv);vector<int> t = cur.vv,nn(6);//Leftnn[0] = t[3],nn[1] = t[2],nn[2] = t[0],nn[3] = t[1],nn[4] = t[4],nn[5] = t[5];q.push({nn,cur.s+1});//Right nn[0] = t[2],nn[1] = t[3],nn[2] = t[1],nn[3] = t[0],nn[4] = t[4],nn[5] = t[5];q.push({nn,cur.s+1});//Frontnn[0] = t[5],nn[1] = t[4],nn[2] = t[2],nn[3] = t[3],nn[4] = t[0],nn[5] = t[1];q.push({nn,cur.s+1});//Backnn[0] = t[4],nn[1] = t[5],nn[2] = t[2],nn[3] = t[3],nn[4] = t[1],nn[5] = t[0];q.push({nn,cur.s+1});}return -1; } int main() {int z,x,c,v,b,n;while(~scanf("%d%d%d%d%d%d",&z,&x,&c,&v,&b,&n)) {aa.clear(); ss.clear();vector<int> a(6);a[0]=z;a[1]=x;a[2]=c;a[3]=v;a[4]=b;a[5]=n;for(int i = 1; i<=6; i++) {scanf("%d",&x);aa.pb(x);}int ans = bfs(a);printf("%d\n",ans);} return 0 ; }

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