HDU-1016 Prime Ring Problem 素数环（DFS）

大意：给出一个n，把前n个数填在这n个圈里形成一个环，使得相邻的圈中的数相加都为素数。

题目：

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 (0 < n < 20).

OutputThe 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

需要注意的点：

1、要记得判断一下最后一个数和第一个数相加是否是素数。

2、因为这道题数据比较小，可以用数组来避免编写判断素数的函数（编了貌似也可以）

下面是代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
int n,a[22]={1},vis[22];
void print(int cur);
int is[40]={0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0};
int main(void)
{
    int k=0;
    while(~scanf("%d",&n))
    {
        printf("Case %d:\n",++k);
        memset(vis,0,sizeof(vis));
        print(1);
        printf("\n");
    }
    return 0;
}
void print(int cur)
{
    int i;
    if(cur==n&&is[a[0]+a[n-1]])
    {
        for(i=0;i<n;i++)
        {
            printf("%d%c",a[i],i==n-1?'\n':' ');
        }
        return;
    }
    else
    {
        for(i=2;i<=n;i++)
        {
            if(is[a[cur-1]+i]&&!vis[i])
            {
                a[cur]=i;
                vis[i]=1;
                print(cur+1);
                vis[i]=0;
            }
        }
    }
}