Prime Ring Problem_DFS

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43088    Accepted Submission(s): 19128

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<cmath>
#include<string.h>
#include<stdio.h>
using namespace std;
const int N=25;
int a[N];
int vis[N],ans[N];
bool prime(int x)
{
    int i;
    for(i=2;i<=sqrt(x);i++)
    {
        if(x%i==0) break;
    }
    if(i>sqrt(x)) return true;
    else return false;
}
void dfs(int k,int n)
{
    if(k==n+1&&prime(ans[n]+ans[1]))//不要忘记他是一个环，首尾也不能相加是素数哦！
    {
        for(int i=1;i<=n;i++)
        {
            if(i!=1)
            cout<<" "<<ans[i];
            else cout<<ans[i];
        }
        cout<<endl;
        return ;
    }

    for(int i=2;i<=n;i++)
    {
        if(prime(ans[k-1]+a[i])&&!vis[a[i]])
        {
            vis[a[i]]=1;
            ans[k]=a[i];
            dfs(k+1,n);
            vis[a[i]]=0;
        }
    }

}
int main()
{
    int n;
    int cas=1;
    while(cin>>n)
    {
        memset(a,0,sizeof(a));
        memset(ans,0,sizeof(ans));
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=n;i++)
        {
            a[i]=i;
        }
        ans[1]=1;vis[1]=1;
        printf("Case %d:
",cas++);
        dfs(2,n);
        printf("
");

    }
    return 0;
}