HDU 4801 Pocket Cube（模拟题——转魔方）
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4801
题面：
Pocket Cube
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 882 Accepted Submission(s): 271
Problem Description
Pocket Cube is a 3-D combination puzzle. It is a 2 × 2 × 2 cube, which means it is constructed by 8 mini-cubes. For a combination of 2 × 2 mini-cubes which sharing a whole cube face, you can twist it 90 degrees in clockwise or counterclockwise direction, this twist operation is called one twist step.
Considering all faces of mini-cubes, there will be totally 24 faces painted in 6 different colors (Indexed from 0), and there will be exactly 4 faces painted in each kind of color. If 4 mini-cubes' faces of same color rely on same large cube face, we can call the large cube face as a completed face.
Now giving you an color arrangement of all 24 faces from a scrambled Pocket Cube, please tell us the maximum possible number of completed faces in no more than N twist steps.
Index of each face is shown as below:
Considering all faces of mini-cubes, there will be totally 24 faces painted in 6 different colors (Indexed from 0), and there will be exactly 4 faces painted in each kind of color. If 4 mini-cubes' faces of same color rely on same large cube face, we can call the large cube face as a completed face.
Now giving you an color arrangement of all 24 faces from a scrambled Pocket Cube, please tell us the maximum possible number of completed faces in no more than N twist steps.
Index of each face is shown as below:
Input
There will be several test cases. In each test case, there will be 2 lines. One integer N (1 ≤ N ≤ 7) in the first line, then 24 integers Ci separated by a single space in the second line. For index 0 ≤ i < 24, Ci is color of the corresponding face. We guarantee that the color arrangement is a valid state which can be achieved by doing a finite number of twist steps from an initial cube whose all 6 large cube faces are completed faces.
Output
For each test case, please output the maximum number of completed faces during no more than N twist step(s).
Sample Input
1 0 0 0 0 1 1 2 2 3 3 1 1 2 2 3 3 4 4 4 4 5 5 5 5 1 0 4 0 4 1 1 2 5 3 3 1 1 2 5 3 3 4 0 4 0 5 2 5 2
Sample Output
6 2
Source
2013 Asia Changsha Regional Contest
解题：
转魔方，注意对应位置别弄错就好，深搜，广搜都可以。就是比较难调。
总结：
一些重复性代码，最好写成函数调用，减少代码的冗余性，同时提高代码的准确性。
代码：
#include <iostream>
#include <cstdio>
using namespace std;
int cube[24],tmp,a,b,c,ans,n;
int check()
{
int res=0;
tmp=cube[0];
if(tmp==cube[1]&&tmp==cube[2]&&tmp==cube[3])res++;
tmp=cube[4];
if(tmp==cube[5]&&tmp==cube[10]&&tmp==cube[11])res++;
tmp=cube[6];
if(tmp==cube[7]&&tmp==cube[12]&&tmp==cube[13])res++;
tmp=cube[8];
if(tmp==cube[9]&&tmp==cube[14]&&tmp==cube[15])res++;
tmp=cube[16];
if(tmp==cube[17]&&tmp==cube[18]&&tmp==cube[19])res++;
tmp=cube[20];
if(tmp==cube[21]&&tmp==cube[22]&&tmp==cube[23])res++;
return res;
}
void rotate(int oper)
{
if(oper==1)
{
a=cube[1];
b=cube[3];
c=cube[9];
cube[1]=cube[7];
cube[3]=cube[13];
cube[7]=cube[17];
cube[13]=cube[19];
cube[17]=cube[21];
cube[19]=cube[23];
cube[21]=a;
cube[23]=b;
cube[9]=cube[8];
cube[8]=cube[14];
cube[14]=cube[15];
cube[15]=c;
}
else if(oper==6)
{
a=cube[7];
b=cube[13];
c=cube[14];
cube[7]=cube[1];
cube[13]=cube[3];
cube[1]=cube[21];
cube[3]=cube[23];
cube[21]=cube[17];
cube[23]=cube[19];
cube[17]=a;
cube[19]=b;
cube[14]=cube[8];
cube[8]=cube[9];
cube[9]=cube[15];
cube[15]=c;
}
else if(oper==2)
{
a=cube[8];
b=cube[14];
c=cube[7];
cube[8]=cube[17];
cube[14]=cube[16];
cube[17]=cube[11];
cube[16]=cube[5];
cube[11]=cube[2];
cube[5]=cube[3];
cube[2]=a;
cube[3]=b;
cube[7]=cube[13];
cube[13]=cube[12];
cube[12]=cube[6];
cube[6]=c;
}
else if(oper==5)
{
a=cube[16];
b=cube[17];
c=cube[12];
cube[17]=cube[8];
cube[16]=cube[14];
cube[8]=cube[2];
cube[14]=cube[3];
cube[2]=cube[11];
cube[3]=cube[5];
cube[11]=b;
cube[5]=a;
cube[12]=cube[13];
cube[13]=cube[7];
cube[7]=cube[6];
cube[6]=c;
}
else if(oper==3)
{
a=cube[12];
b=cube[13];
c=cube[16];
cube[12]=cube[14];
cube[13]=cube[15];
cube[14]=cube[21];
cube[15]=cube[20];
cube[21]=cube[10];
cube[20]=cube[11];
cube[10]=a;
cube[11]=b;
cube[16]=cube[17];
cube[17]=cube[19];
cube[19]=cube[18];
cube[18]=c;
}
else
{
a=cube[12];
b=cube[13];
c=cube[16];
cube[12]=cube[10];
cube[13]=cube[11];
cube[10]=cube[21];
cube[11]=cube[20];
cube[20]=cube[15];
cube[21]=cube[14];
cube[14]=a;
cube[15]=b;
cube[16]=cube[18];
cube[18]=cube[19];
cube[19]=cube[17];
cube[17]=c;
}
}
void search(int oper,int step)
{
rotate(oper);
tmp=check();
if(tmp>ans)ans=tmp;
if(step==n)
{
rotate(7-oper);
return;
}
for(int i=1;i<=6;i++)
{
if(i!=(7-oper))
{
search(i,step+1);
}
}
rotate(7-oper);
}
int main()
{
while(~scanf("%d",&n))
{
for(int i=0;i<24;i++)
scanf("%d",&cube[i]);
ans=check();
for(int i=1;i<=6;i++)
search(i,1);
printf("%d\n",ans);
}
return 0;
}