Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34609    Accepted Submission(s): 15327

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

这道题就是回溯暴力。 首先打出一个素数表， 然后DFS回溯判断即可。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int is_prime(int x)
{
    for(int i=2; i*i<=x; i++)
    if(x%i==0) return 0;
    return 1;
}

int n, A[50];
bool isp[50], vis[50];
void dfs(int cur)
{
    if(cur==n&&isp[A[0]+A[n-1]])
    {
        for(int i=0; i<n-1; i++)
        {
            printf("%d ", A[i]);
        }
        printf("%d\n", A[n-1]);
    }
    else for(int i=2; i<=n; i++)
    if(!vis[i]&&isp[i+A[cur-1]])
    {
        A[cur] = i;
        vis[i] = 1;
        dfs(cur+1);
        vis[i]=0;
    }
}

int main()
{
    int kase = 0;
    while(scanf("%d", &n)!=EOF)
    {
        printf("Case %d:\n", ++kase);
        for(int i=2; i<=n*2; i++)
        isp[i] = is_prime(i);
        memset(vis, 0, sizeof(vis));
        A[0] = 1;
        dfs(1);
        printf("\n");
    }
    return 0;
}