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).
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++) {