Another Eight Puzzle

Problem Description

Fill the
following 8 circles with digits 1~8,with each number exactly once .
Conntcted circles cannot be filled with two consecutive
numbers.

There are 17 pairs of connected cicles:

A-B , A-C, A-D

B-C, B-E, B-F

C-D, C-E, C-F, C-G

D-F, D-G

E-F, E-H

F-G, F-H

G-H

Eight Puzzle" title="Another Eight Puzzle">


Filling G with 1 and D with 2 (or G with 2 and D with 1) is illegal
since G and D are connected and 1 and 2 are consecutive .However
,filling A with 8 and B with 1 is legal since 8 and 1 are not
consecutive .



In this problems,some circles are already filled,your tast is to
fill the remaining circles to obtain a solution (if
possivle).

Input

The first line
contains a single integer T(1≤T≤10),the number of test cases. Each
test case is a single line containing 8 integers 0~8,the numbers in
circle A~H.0 indicates an empty circle.

Output

For each test
case ,print the case number and the solution in the same format as
the input . if there is no solution ,print “No answer”.If there
more than one solution,print “Not unique”.

Sample Input

3

7 3 1 4 5 8
0 0

7 0 0 0 0 0
0 0

1 0 0 0 0 0
0 0

Sample Output

Case 1: 7 3
1 4 5 8 6 2

Case 2: Not
unique

Case 3: No
answer

题意：如图所示，有A-H 8个圆圈，关系如图所示，给出各个圆圈内的数，0代表你要填的数，让你求要多少种填法；

解题思路：刚看到时，以为和相邻两个数的和是素数那个差不多，后来写的时候，有点不一样那个，从第一位开始搜索，位上有数字就跳过，如果没有就搜索：

A-B , A-C,
A-D

   B-A, B-C,
B-E, B-F

   C-A, C-B, C-D, C-F,
C-E,C-G

   D-A, D-C, D-F, D-G

   E-B, E-C, E-F, E-H

   F-C, F-D, F-G, F-H, F-E,
F-B

   G-D, G-C, G-F, G-H

   H-E, H-F, H-G

   A=1 B=2 C=3 D=4 E=5 F=6 G=7
H=8

每到一位的判断关系；

感悟：判断条件老是写不对，真是费劲啊

代码：

#include

#include

#include

#include

#include

#define maxn 10

using namespace std;

int pos[maxn],t,visit[maxn],ans=0,mark[maxn],p[maxn];



bool conect(int x,int y)

{

    if (x>y)
{int t=x; x=y; y=t;}

    switch
(x)

    {

       
case 1:return y==2 || y==3 || y==4;

       
case 2:return y==3 || y==5 || y==6;

       
case 3:return y==4 || y==5 || y==6 || y==7;

       
case 4:return y==6 || y==7;

       
case 5:return y==6 || y==8;

       
case 6:return y==7 || y==8;

       
case 7:return y==8;

    }

}

bool ok(int p)

{

    int i;

    for
(i=1;i<=8;i++)

       
if ((conect(i,p) && abs(pos[p]-pos[i])==1) &&
pos[i]!=0) return 0;

    return
1;

}



void dfs(int cur)

{

   
//cout<<cur<<" ";

   
if(ans>1)

       
return ;//剪枝大于2了就不需要在搜了

   
//cout<<"cur="<<cur<<endl;

   
while(cur<=8&&pos[cur]) cur++;//把0空过去

   
if(cur>8)

    {

       
ans++;

       
if(ans==1)//因为只有ans=1时才需要输出

           
memcpy(p,pos,sizeof(pos));//将整个数组转移过来

       
return;

    }

    for(int
i=1;i<=8;i++) if(!visit[i])

    {

       
//cout<<cur<<endl;

       
//cout<<i<<" ";

       
visit[i]=1;

       
pos[cur]=i;

       
if(ok(cur))//相邻的不是相邻的数

           
dfs(cur+1);

       
pos[cur]=0;//原来这里就是零的

       
visit[i]=0;//释放标记，用于下一次搜索；

       
if(ans>1)

           
return;

    }

}

int main()

{

   
//freopen("in.txt", "r", stdin);

    int
flag=0;

   
scanf("%d",&t);

    for(int
i=1;i<=t;i++)

    {

       
ans=0;

       
memset(pos,0,sizeof(pos));

       
memset(visit,0,sizeof(visit));

       
for(int j=1;j<=8;j++)

       
{

           
scanf("%d",&pos[j]);

           
visit[pos[j]]=1;

       
}

       
dfs(1);

       
//cout<<"ans="<<ans<<endl;

       
printf("Case %d: ",i);

       
if(!ans)

           
printf("No answer");

       
else if(ans==1)

           
for(int i=1;i<=8;i++)

               
printf(i==1?"%d":" %d",p[i]);

       
else

           
printf("Not unique");

       
printf("\n");

    }

}