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

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(" ");
    }
}