标签：des style blog http color io os ar for
Problem Description
There are 2 special dices on the table. On each face of the dice, a distinct number was written. Consider a1.a2,a3,a4,a5,a6
to be numbers written on top face, bottom face, left face, right face,
front face and back face of dice A. Similarly, consider b1.b2,b3,b4,b5,b6
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 ai ≠ aj and bi ≠ bj 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, 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)
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 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.
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
这个题目可以用bfs遍历向前、向后、向左、向右转 ，这样如果用一个数组a[6]记录一种状态，那么最多也只有6！种状态，数量不是很多，可以直接暴力bfs。不过需要记录每个状态是否被访问过。
代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cstring> 5 #include <cmath> 6 #include <algorithm> 7 #include <set> 8 #include <map> 9 #include <queue> 10 #include <string> 11 #include <vector> 12 #define inf 0x3fffffff 13 #define esp 1e-10 14 using namespace std; 15 struct node1 16 { 17 int dice[6]; 18 int val; 19 }; 20 struct node 21 { 22 node1 qt; 23 int step; 24 }; 25 node a; 26 node1 b; 27 int bfs() 28 { 29 set < int > s; 30 s.insert(a.qt.val); 31 queue < node > q; 32 q.push(a); 33 while (!q.empty()) 34 { 35 node f, k; 36 f = q.front(); 37 q.pop(); 38 if (f.qt.val == b.val) return f.step; 39 //first 40 k = f; 41 swap (k.qt.dice[0], k.qt.dice[5]); 42 swap (k.qt.dice[4], k.qt.dice[1]); 43 swap (k.qt.dice[5], k.qt.dice[4]); 44 k.qt.val = k.qt.dice[0]; 45 for (int y = 1; y < 6; ++y) 46 { 47 k.qt.val = 10*k.qt.val + k.qt.dice[y]; 48 } 49 if (s.find(k.qt.val) == s.end()) 50 { 51 k.step ++; 52 q.push(k); 53 s.insert(k.qt.val); 54 k.step --; 55 } 56 //second 57 k = f; 58 swap (k.qt.dice[0], k.qt.dice[5]); 59 swap (k.qt.dice[4], k.qt.dice[1]); 60 swap (k.qt.dice[0], k.qt.dice[1]); 61 k.qt.val = k.qt.dice[0]; 62 for (int y = 1; y < 6; ++y) 63 { 64 k.qt.val = 10*k.qt.val + k.qt.dice[y]; 65 } 66 if (s.find(k.qt.val) == s.end()) 67 { 68 k.step ++; 69 q.push(k); 70 s.insert(k.qt.val); 71 k.step --; 72 } 73 //third 74 k = f; 75 swap (k.qt.dice[0], k.qt.dice[2]); 76 swap (k.qt.dice[1], k.qt.dice[3]); 77 swap (k.qt.dice[1], k.qt.dice[0]); 78 k.qt.val = k.qt.dice[0]; 79 for (int y = 1; y < 6; ++y) 80 { 81 k.qt.val = 10*k.qt.val + k.qt.dice[y]; 82 } 83 if (s.find(k.qt.val) == s.end()) 84 { 85 k.step ++; 86 q.push(k); 87 s.insert(k.qt.val); 88 k.step --; 89 } 90 //forth 91 k = f; 92 swap (k.qt.dice[0], k.qt.dice[2]); 93 swap (k.qt.dice[1], k.qt.dice[3]); 94 swap (k.qt.dice[2], k.qt.dice[3]); 95 k.qt.val = k.qt.dice[0]; 96 for (int y = 1; y < 6; ++y) 97 { 98 k.qt.val = 10*k.qt.val + k.qt.dice[y]; 99 } 100 if (s.find(k.qt.val) == s.end()) 101 { 102 k.step ++; 103 q.push(k); 104 s.insert(k.qt.val); 105 k.step --; 106 } 107 } 108 return -1; 109 } 110 int main() 111 { 112 //freopen ("test.txt", "r", stdin); 113 while (scanf ("%d", &a.qt.dice[0]) != EOF) 114 { 115 for (int i = 1; i < 6; ++i) 116 scanf ("%d", &a.qt.dice[i]); 117 a.step = 0; 118 a.qt.val = a.qt.dice[0]; 119 for (int y = 1; y < 6; ++y) 120 { 121 a.qt.val = 10*a.qt.val + a.qt.dice[y]; 122 } 123 for (int i = 0; i < 6; ++i) 124 scanf ("%d", &b.dice[i]); 125 b.val = b.dice[0]; 126 for (int y = 1; y < 6; ++y) 127 { 128 b.val = 10*b.val + b.dice[y]; 129 } 130 printf ("%d\n", bfs()); 131 } 132 return 0; 133 }
ACM学习历程—HDU 5012 Dice（ACM西安网赛）（bfs）
标签：des style blog http color io os ar for
原文地址：http://www.cnblogs.com/andyqsmart/p/4014507.html
