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.
For each case, the first line consists of six integers a 1<??http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">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"brush:java;">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 题意：求起始到终态的步骤 思路：简单的BFS+判重，注意判重数组开准点#include#include #include #include #include using namespace std; const int maxn = 6; struct Node{ int arr[maxn], step; Node() { memset(arr, 0, sizeof(arr)); step = 0; } }start, end; int vis[maxn*200000]; int cal(Node a) { int num = 0; for (int i = 0; i < maxn; i++) { num = num * 10 + a.arr[i]; } return num; } bool equal(Node a, Node b) { for (int i = 0; i < maxn; i++) if (a.arr[i] != b.arr[i]) return false; return true; } Node turn(Node a, int i) { Node c; if (i == 1) { c.arr[0] = a.arr[3]; c.arr[1] = a.arr[2]; c.arr[2] = a.arr[0]; c.arr[3] = a.arr[1]; c.arr[4] = a.arr[4]; c.arr[5] = a.arr[5]; } if (i == 2) { c.arr[0] = a.arr[2]; c.arr[1] = a.arr[3]; c.arr[2] = a.arr[1]; c.arr[3] = a.arr[0]; c.arr[4] = a.arr[4]; c.arr[5] = a.arr[5]; } if (i == 3) { c.arr[0] = a.arr[5]; c.arr[1] = a.arr[4]; c.arr[2] = a.arr[2]; c.arr[3] = a.arr[3]; c.arr[4] = a.arr[0]; c.arr[5] = a.arr[1]; } if (i == 4) { c.arr[0] = a.arr[4]; c.arr[1] = a.arr[5]; c.arr[2] = a.arr[2]; c.arr[3] = a.arr[3]; c.arr[4] = a.arr[1]; c.arr[5] = a.arr[0]; } return c; } int bfs() { memset(vis, 0, sizeof(vis)); queue q; q.push(start); Node tmp; vis[cal(start)] = 1; while (!q.empty()) { tmp = q.front(); q.pop(); if (equal(tmp, end)) { return tmp.step; } for (int i = 1; i <= 4; i++) { Node c; c = turn(tmp, i); if (!vis[cal(c)]) { c.step = tmp.step + 1; vis[cal(c)] = 1; q.push(c); } } } return -1; } int main() { while (scanf("%d", &start.arr[0]) != EOF) { for (int i = 1; i < maxn; i++) scanf("%d", &start.arr[i]); for (int i = 0; i < maxn; i++) scanf("%d", &end.arr[i]); printf("%d\n", bfs()); } return 0; }