2023

2023

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.<br>There are 17 pairs of connected cicles:<br>A-B , A-C, A-D<br>B-C, B-E, B-F<br>C-D, C-E, C-F, C-G<br>D-F, D-G<br>E-F, E-H<br>F-G, F-H<br>G-H<br><center><img src=../../../data/images/C150-1010-1.jpg></center><br>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 .<br><br>In this problems,some circles are already filled,your tast is to fill the remaining circles to obtain a solution (if possivle).<br>

 

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.<br><br>

 

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”.<br>

 

Sample Input

   
   
   
   

    
    
    
    3<br>7 3 1 4 5 8 0 0<br>7 0 0 0 0 0 0 0<br>1 0 0 0 0 0 0 0<br>
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case 1: 7 3 1 4 5 8 6 2<br>Case 2: Not unique<br>Case 3: No answer<br>
   
   
   
   

 

Source

ECJTU 2008 Autumn Contest

 

题意：

输入8个数 表示 图中a b c d e f g h的位置的值只能为1-8的数 然后相邻的不能是连续的数 即绝对值不能为1  有些数是0 将为0的填成1-8中未使用的数

有多少种方法  如果仅有一种 输出它   有多种 或没有  按样例中那样输出  

思路：这题感觉跟数独那道题有点像。很显然这题用深搜很合适。但是以前有一个题目，我没有考虑到其他的，直接搜索，导致超出时间限制，在这道题里面，对于是否符合条件的限定，我先列出来了。这样应该可以省下不少的时间；

AC代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
int a[10],vis[10],ans[10],anscnt;
int abs(int q)
{
if(q<0) return -q;
return q;
}
int ok()
{
if(abs(a[2]-a[1])!=1&&
abs(a[3]-a[1])!=1&&
abs(a[4]-a[1])!=1&&


abs(a[2]-a[3])!=1&&
abs(a[2]-a[5])!=1&&
abs(a[2]-a[6])!=1&&


abs(a[3]-a[4])!=1&&
abs(a[3]-a[5])!=1&&
abs(a[3]-a[6])!=1&&
abs(a[3]-a[7])!=1&&


abs(a[4]-a[6])!=1&&
abs(a[4]-a[7])!=1&&


abs(a[5]-a[6])!=1&&
abs(a[5]-a[8])!=1&&


abs(a[6]-a[7])!=1&&
abs(a[6]-a[8])!=1&&


abs(a[7]-a[8])!=1
)
return 1;
else return 0;
}
void DFS(int k)
{
    int i,cnt=0;
if(k==9)
//注意这里是k==9 而不是把让它等于8之后放在调用函数的最后面 那样的话 最后一次的赋值就会被还原为0
{
if(ok())
{
anscnt++;
if(anscnt==1)
{
for(i=1;i<=8;i++)
ans[i]=a[i];
}
}
return;
}
    if(anscnt>=2) return;
    if(a[k]!=0) DFS(k+1);
else
for(i=1;i<=8;i++)
{
if(!vis[i])
{
a[k]=i;
vis[i]=1;
DFS(k+1);
a[k]=0;
vis[i]=0;
}
}
}


int main()
{
int t,cas=0;
scanf("%d",&t);
while(t--)
{
anscnt=0;
int i;
memset(vis,0,sizeof(vis));
memset(ans,0,sizeof(ans));
scanf("%d %d %d %d %d %d %d %d",&a[1],&a[2],&a[3],&a[4],&a[5],&a[6],&a[7],&a[8]);
for(i=1;i<=8;i++) vis[a[i]]=1;
        DFS(1);
        printf("Case %d: ",++cas);
        if(anscnt==1)
        {
for(i=1;i<8;i++)  printf("%d ",ans[i]);
printf("%d\n",ans[i]);
        }
        else if(anscnt==0) printf("No answer\n");
        else printf("Not unique\n");
}
return 0;
}