HUAS Summer Trainning #3 C

Description

 

A ring is composed of n (even number) circles as shown in diagram. Put natural numbers  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 <= 16)

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.

You are to write a program that completes above process.

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

题目大意：使1~n这些数组成环，并且两两相加要是素数，输出所有的解。

解题思路:第一个数总是1，直接用DFS从第二数开始遍历，遍历到底部不满足，回溯到上一个个节点，知道遍历完所有的可行解。

代码：

#include<iostream>
#include<cstring>
using namespace std;
const int maxn=50+5;
int a[maxn],b[maxn],d[maxn],n,i,j;
int sushu(int x)
{
    for(i=2;i*i<=x;i++)
        if(x%i==0)
            return 0;
    return 1;
}
void dfs(int cur)
{
    if(cur==n&&d[a[0]+a[cur-1]])
    {
        cout<<a[0];
        for(int  i=1;i<n;i++)
            cout<<" "<<a[i];
        cout<<endl;
    }
    else
    {
        for(int i=2;i<=n;i++)
        {
            if(!b[i]&&d[i+a[cur-1]])
            {
                a[cur]=i;
                b[i]=1;
                dfs(cur+1);
                b[i]=0;
            }
        }
    }
}
int main()
{
    int p=0,t=0;
    while(cin>>n&&n)
    {
        if(p++)
            cout<<endl;    
        cout<<"Case "<<++t<<":"<<endl;
    
      memset(a,0,sizeof(a));
      memset(d,0,sizeof(d));
      for(j=2;j<n*2;j++)
          d[j]=sushu(j);
      memset(b,0,sizeof(b));
      a[0]=1;
      dfs(1);
    }
    return 0;
}