SDAU 搜索专题 23 Another Eight Puzzle

1：问题描述

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

2：大致思路



将数字填到表中要使有连线的位置，不能相邻。

3：思路

枚举就好啦。先将表填满，再检索是否满足题意。

4：感想

心好累。〒_〒

5：ac代码

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<iostream>
#include<cstdio>
using namespace std;
int a[10],vis[10],temp[10],cou;
int isok()
{
if(fabs(a[2]-a[1])!=1&&
fabs(a[3]-a[1])!=1&&
fabs(a[4]-a[1])!=1&&

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

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

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

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

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

fabs(a[7]-a[8])!=1
)
return 1;
else return 0;
}
void dfs(int num)
{
int i;
if(num==9)
{
if(isok())
{
cou++;
if(cou==1)
{
for(i=1;i<=8;i++)
temp[i]=a[i];
}
}
return;
}
if(cou>=2) return;
if(a[num]==0)
for(i=1;i<=8;i++)
{
if(!vis[i])
{
a[num]=i;
vis[i]=1;
dfs(num+1);
a[num]=0;
vis[i]=0;
}
}
else dfs(num+1);
}

int main()
{
//freopen("r.txt","r",stdin);
int t,casee=0;
scanf("%d",&t);
while(t--)
{
casee++;
cou=0;
int i;
memset(vis,0,sizeof(vis));
memset(temp,0,sizeof(temp));
for(i=1;i<=8;i++)
{
scanf("%d",&a[i]);//注意这里有个空格哦！
vis[a[i]]=1;
}
for(i=1;i<=8;i++)
if(a[i]==0)
break;
dfs(i);
cout<<"Case "<<casee<<": ";
if(cou==1)
{
cout<<temp[1];
for(i=2;i<=8;i++)
cout<<" "<<temp[i];
cout<<endl;
}
else if(cou==2)
cout<<"Not unique"<<endl;
else
cout<<"No answer"<<endl;
}
return 0;
}