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
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).
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
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++)
{
#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)
{
}
bool ok(int p)
{
}
void dfs(int cur)
{
}
int main()
{