hdu 1016 Prime Ring Problem

http://acm.hdu.edu.cn/showproblem.php?pid=1016

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19551    Accepted Submission(s): 8742

Problem Description

A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime.

Note: the number of first circle should always be 1.

 

Input

n (0 < n < 20).

 

Output

The output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements. Print solutions in lexicographical order.

You are to write a program that completes above process.

Print a blank line after each case.

 

Sample Input

6 8

 

Sample Output

Case 1:

1 4 3 2 5 6

1 6 5 2 3 4

 

Case 2:

1 2 3 8 5 6 7 4

1 2 5 8 3 4 7 6

1 4 7 6 5 8 3 2

1 6 7 4 3 8 5 2

 

Source

Asia 1996, Shanghai (Mainland China)

 

Recommend

JGShining

 就一个dfs

#include<stdio.h>
#include<string.h>
#include<math.h>
int num[20],vis[20],n;
int is_prime(int x)
{
     int i;
     for(i=2;i<=sqrt(x);i++)
     if(x%i==0)
     return 0;
     return 1;

}
void dfs(int x,int k)
{
     int i;
     if(k==n &&is_prime(1+x))
     {
          for(i=0;i<n;i++)
          printf("%d%c",num[i],i==k-1?'
':' ');
     }
     for(i=2;i<=n;i++)
     {
          if(is_prime(x+i)&&!vis[i])
          {
            // printf("x=%d,i=%d,k=%d
",x,i,k);
               vis[i]=1;
               num[k]=i;
               dfs(i,k+1);
               vis[i]=0;
          }
     }
}
int main()
{
    int count=1;
     while(scanf("%d",&n)!=EOF)
     {
          memset(vis,0,sizeof(vis));
          printf("Case %d:
",count++);
          num[0]=1;
          dfs(1,1);
          printf("
");

     }
     return 0;
}