hdu1016 Prime Ring Problem(DFS)

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 66391    Accepted Submission(s): 28456

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

 

//HD1016
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
static int j = ;
int n, a[];
bool prime[], vis[];

bool is_prime(int m)//素数判断
{
    if (m == )//1不是素数
        return false;
    for (int i = ; i*i <= m; i++)//素数只能被1和本身整除
        if (m % i == )
            return false;
    return true;
}
void dfs(int cur)
{
    if (cur == n && prime[a[] + a[n - ]])//递归出口，当到最后一个数并且首位相加为素数就结束递归
    {
        for (int i = ; i < n; i++)
        {
            cout << a[i];
            if (i < n - )
                cout << ' ';
            else
                cout << endl;
        }
    }
    else
        for (int i = ; i <= n; i++)
        if (!vis[i] && prime[i + a[cur - ]])
        {
            a[cur] = i;
            vis[i] = ;
            dfs(cur + );
            vis[i] = ;
        }
}

int main()
{
    memset(a, , sizeof(a));
    memset(prime, false, sizeof(prime));
    memset(vis, false, sizeof(vis));
    for (int i = ; i < ; i++)
        prime[i] = is_prime(i);
    while (scanf("%d", &n) != EOF)
    {
        a[] = ;
        printf("Case %d:\n", j++);
        dfs();
        cout << endl;
    }
    return ;
}