ZOJ 1457 Prime Ring Problem（dfs+剪枝）

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<cstdio>
#include<cstring>
#include<cmath>

using namespace std;

int n,test=0;;
int cir[30];
int vis[30];
int prime[50]={0};

void dfs(int k)
{
    if(k==n)
    {
        if(!prime[cir[k]+cir[1]])
            return;
        for(int i=1;i<n;i++)
           printf("%d ",cir[i]);
        printf("%d",cir[n]);
        printf("\n");
        return;
    }

    for(int j=2;j<=n;j++)
    {
        if(!vis[j]&&prime[cir[k]+j])       //对于相邻素数的推断，在这里才干在zoj,AC，放最前面就仅仅能在HDU。AC了，一层递归调用的时间而已啊！
        {
      
            vis[j]=1;
            cir[k+1]=j;
            dfs(k+1);
            vis[j]=0;                      //注意递归复原，这预计是最大的亮点了

        }

    }
}

int main()
{
    prime[3]=1;
    prime[5]=1;
    prime[7]=1;
    prime[11]=1;
    prime[13]=1;
    prime[17]=1;
    prime[19]=1;
    prime[23]=1;
    prime[29]=1;
    prime[31]=1;
    prime[37]=1;


    while(scanf("%d",&n)!=EOF)
    {
        //memset(cir,0,sizeof cir);
        //memset(vis,0,sizeof vis);
        for(int i=1;i<=n;i++)
        {
            vis[i]=0;
        }
        printf("Case %d:\n",++test);
        if(n % 2 == 1)
        {
            printf("\n");
            continue;
        }

        cir[1]=1;
        dfs(1);
        printf("\n");
    }


    return 0;
}