Another Eight Puzzle （枚举深搜）

hdu2514 Another Eight Puzzle 填数字 搜索水题
2013-05-21 08:43:08     我来说两句       作者：hnust_xiehonghao
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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

[cpp] #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; 

}