HDU5012 Dice

Dice

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 316 Accepted Submission(s): 188

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

Source

2014 ACM/ICPC Asia Regional Xi'an Online

题意：两个色子，给出两个色子的初始状态（上下左右前后），问前面一个能不能通过转动变成后面一个。

写出四种转法转动后的状态，然后直接搜索。。。。记忆化搜索和BFS都行。。。好像我写的记忆化比BFS快....

记忆化搜索

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int INF=1<<30;
int dp[7][7][7][7][7][7];
int su,sd,sl,sr,sf,sre;
int fu,fd,fl,fr,ff,fre;
int flag;
int dfs(int up,int down,int left,int right,int front,int rear)
{
if(dp[up][down][left][right][front][rear]!=-1)
return dp[up][down][left][right][front][rear];
if(up==fu&&down==fd&&left==fl&&right==fr&&front==ff&&rear==fre)
{
flag=1;
return dp[up][down][left][right][front][rear]=0;
}
dp[up][down][left][right][front][rear]=INF;
int temp=INF;
temp=min(temp,dfs(rear,front,left,right,up,down));
temp=min(temp,dfs(front,rear,left,right,down,up));
temp=min(temp,dfs(right,left,up,down,front,rear));
temp=min(temp,dfs(left,right,down,up,front,rear));
return dp[up][down][left][right][front][rear]=temp+1;
}
int main()
{
while(scanf("%d%d%d%d%d%d",&su,&sd,&sl,&sr,&sf,&sre)==6)
{
scanf("%d%d%d%d%d%d",&fu,&fd,&fl,&fr,&ff,&fre);
memset(dp,-1,sizeof(dp));
flag=0;
int ans=dfs(su,sd,sl,sr,sf,sre);
if(flag)
printf("%d\n",ans);
else
printf("-1\n");
}
return 0;
}

BFS

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
struct Node
{
int a[6],d;
Node()
{
memset(a,0,sizeof(a));
d=0;
}
}s,e;
int hash(Node x)
{
int num=0;
for(int i=0;i<6;i++)
num=num*10+x.a[i];
return num;
}
Node turn(Node x,int i)
{
Node c;
if(i==1)
{
c.a[0]=x.a[5];
c.a[1]=x.a[4];
c.a[2]=x.a[2];
c.a[3]=x.a[3];
c.a[4]=x.a[0];
c.a[5]=x.a[1];
}
else if(i==2)
{
c.a[0]=x.a[4];
c.a[1]=x.a[5];
c.a[2]=x.a[2];
c.a[3]=x.a[3];
c.a[4]=x.a[1];
c.a[5]=x.a[0];
}
else if(i==3)
{
c.a[0]=x.a[3];
c.a[1]=x.a[2];
c.a[2]=x.a[0];
c.a[3]=x.a[1];
c.a[4]=x.a[4];
c.a[5]=x.a[5];
}
else if(i==4)
{
c.a[0]=x.a[2];
c.a[1]=x.a[3];
c.a[2]=x.a[1];
c.a[3]=x.a[0];
c.a[4]=x.a[4];
c.a[5]=x.a[5];
}
return c;
}
bool judge(Node x)
{
for(int i=0;i<6;i++)
if(x.a[i]!=e.a[i])
return 0;
return 1;
}
bool vis[2000010];
int bfs()
{
memset(vis,0,sizeof(vis));
queue<Node> q;
vis[hash(s)]=1;
q.push(s);
while(!q.empty())
{
Node u=q.front();
q.pop();
if(judge(u))
return u.d;
for(int i=1;i<=4;i++)
{
Node v=turn(u,i);
if(!vis[hash(v)])
{
v.d=u.d+1;
q.push(v);
vis[hash(v)]=1;
}
}
}
return -1;
}
int main()
{
int i;
while(scanf("%d",&s.a[0])==1)
{
for(i=1;i<6;i++)
scanf("%d",&s.a[i]);
for(i=0;i<6;i++)
scanf("%d",&e.a[i]);
printf("%d\n",bfs());
}
return 0;
}