Prime Ring Problem HDU

Prime Ring Problem

 HDU - 1016 

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. 

 

Inputn (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

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<queue>

using namespace std;


int n;
int a[23];
int vis[23];
bool mark[1000];

void init()        // 素数筛 
{
    for(int i = 2; i < 1000; ++i)
        mark[i] = true;
    
    for(int i = 2; i < 1000; ++i)
    {
        if(mark[i] == true)
        {
            for(int j = i*i; j < 1000; j += i)
            {
                mark[j] = false;
            }
        }
    }
}

// 边枚举边判断，不要最后一次性判断，会超时 
void dfs(int step)
{
    if(step > 2)
    {
        if(mark[a[step-1]+a[step-2]] == false)    // 判断最后两个数 
            return;
    }
    
    if(step == n+1)
    {
        if(mark[a[n]+a[1]] == false)    // 判断最后一个数与第一个数 
            return;
        for(int i = 1; i < n; ++i)
            printf("%d ", a[i]);
        printf("%d
", a[n]);

    }
            
    for(int i = 2; i <= n; ++i)
    {
        if(!vis[i])
        {
            a[step] = i;
            vis[i] = 1;
            dfs(step+1);
            vis[i] = 0;
        }
        
    }
}


int main()
{
     init();
     int cas = 1;
     a[1] = 1;
     while(scanf("%d", &n) != EOF)
     {
         printf("Case %d:
", cas++);
         memset(vis, 0, sizeof(vis));
         dfs(2);
         printf("
");
     }
    
    
    return 0;
}