题目链接

题意：给你两个色子，如果可以通过旋转使两个色子的面一一对应相等(旋转规则题目给出)，求最小步数，如果不行，输出-1。

思路：用BFS求最少步数。

代码：

#include 
  
   
#include 
   
     #include 
    
      #include 
     
       #include 
      
        using namespace std; const int MAXN = 6; struct node{ node() { memset(arr, 0, sizeof(arr)); d = 0; } int arr[MAXN], d; }s, e; int vis[MAXN * 200000]; int change(node a) { int num = 0; for (int i = 0; i < MAXN; i++) { num = num * 10 + a.arr[i]; } return num; } bool judge(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(s); node tmp; vis[change(s)] = 1; while (!q.empty()) { tmp = q.front(); q.pop(); if (judge(tmp, e)) { return tmp.d; } for (int i = 1; i <= 4; i++) { node c; c = turn(tmp, i); if (!vis[change(c)]) { c.d = tmp.d + 1; vis[change(c)] = 1; q.push(c); } } } return -1; } int main() { while (scanf("%d", &s.arr[0]) != EOF) { for (int i = 1; i < MAXN; i++) scanf("%d", &s.arr[i]); for (int i = 0; i < MAXN; i++) scanf("%d", &e.arr[i]); printf("%d\n", bfs()); } return 0; }