素数环 dfs+回溯

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 73040    Accepted Submission(s): 31109

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

 

#include<iostream> 
#include<stack>
#include<queue>
#include<cstring>
#define maxn 1000
using namespace std;
int a[maxn]={0},b[maxn]={0};
int check(int n,int m)
{
    n=n+m;
    for(int i=2;i*i<=n;i++)
      {
           if(n%i==0)
           return false;
      }
      return true;
}
void dfs( int step,int n)
{
    if(step>n)
    return ;
    if(step==n&&(check(a[n-1],a[0])))
    {
    for(int i=0;i<n;i++)
    {
        if(i!=n-1)
        cout<<a[i]<<" ";
        else
        cout<<a[i];
    }
    cout<<endl;
    }
    for(int i=2;i<=n;i++)
    {
      if(check(i,a[step-1])&&b[i]==0)    
      {
          a[step]=i;
          b[i]=1;
          dfs(step+1,n);
          b[i]=0;
      }
    }
}
int main()
{
    int n,k=1;
    while(cin>>n)
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        a[0]=1;
        b[0]=1;
        cout<<"Case "<<k<<":"<<endl;
        dfs(1,n);
        cout<<endl;
        k++;
    }
    return 0;
}