hdu2514 Another Eight Puzzle 填数字 搜索水题

Another Eight Puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 659    Accepted Submission(s): 405

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
   
   
   
   

 

Source

ECJTU 2008 Autumn Contest

 

Recommend

lcy

 

http://acm.hdu.edu.cn/showproblem.php?pid=2514

 

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 DFS 搜

#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;
}