给你一个正方体的初始状态和末状态,问你是否可以再6步之内转到这个状态,有四种转的方式,如果你面对的是正方向的正前方,那么转的方式就是 顺时针,逆时针,上,下。
思路:
水搜索,直接搜就行,深搜广搜随意,mark不mark也随意,因为状态4,深度6那么最多也就4^6,题目没有坑点,细心点就行了,我写的是一个广搜(mark了)。
#include<stdio.h> #include<queue> using namespace std; typedef struct { int now[7]; }NODE; NODE xin ,tou; int mark[7][7][7][7][7][7]; int ss[7] ,ee[7]; int BFS() { memset(mark ,0 ,sizeof(mark)); mark[ss[1]][ss[2]][ss[3]][ss[4]][ss[5]][ss[6]] = 1; queue<NODE>q; for(int i = 0 ;i <= 6 ;i ++) xin.now[i] = ss[i]; q.push(xin); while(!q.empty()) { tou = q.front(); q.pop(); int ok = 1; for(int i = 1 ;i <= 6 ;i ++) if(tou.now[i] != ee[i]) ok = 0; if(ok) return tou.now[0]; if(tou.now[0] >= 6) continue; for(int i = 0 ;i <= 6 ;i ++) xin.now[i] = tou.now[i]; xin.now[0] ++; xin.now[1] = tou.now[3] ,xin.now[3] = tou.now[2] ,xin.now[2] = tou.now[4] ,xin.now[4] = tou.now[1]; if(!mark[xin.now[1]][xin.now[2]][xin.now[3]][xin.now[4]][xin.now[5]][xin.now[6]] ) q.push(xin); for(int i = 0 ;i <= 6 ;i ++) xin.now[i] = tou.now[i]; xin.now[0] ++; xin.now[1] = tou.now[4] ,xin.now[4] = tou.now[2] ,xin.now[2] = tou.now[3] ,xin.now[3] = tou.now[1]; if(!mark[xin.now[1]][xin.now[2]][xin.now[3]][xin.now[4]][xin.now[5]][xin.now[6]] ) q.push(xin); for(int i = 0 ;i <= 6 ;i ++) xin.now[i] = tou.now[i]; xin.now[0] ++; xin.now[1] = tou.now[5] ,xin.now[5] = tou.now[2] ,xin.now[2] = tou.now[6] ,xin.now[6] = tou.now[1]; if(!mark[xin.now[1]][xin.now[2]][xin.now[3]][xin.now[4]][xin.now[5]][xin.now[6]] ) q.push(xin); for(int i = 0 ;i <= 6 ;i ++) xin.now[i] = tou.now[i]; xin.now[0] ++; xin.now[1] = tou.now[6] ,xin.now[6] = tou.now[2] ,xin.now[2] = tou.now[5] ,xin.now[5] = tou.now[1]; if(!mark[xin.now[1]][xin.now[2]][xin.now[3]][xin.now[4]][xin.now[5]][xin.now[6]] ) q.push(xin); } return -1; } int main () { while(~scanf("%d %d %d %d %d %d" ,&ss[1] ,&ss[2] ,&ss[3] ,&ss[4] ,&ss[5] ,&ss[6])) { for(int i = 1 ;i <= 6 ;i ++) scanf("%d" ,&ee[i]); ss[0] = ee[0] = 0; printf("%d\n" ,BFS()); } return 0; }