题意:
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;
}