Prime Ring Problem + DFS

Prime Ring Problem

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 9   Accepted Submission(s) : 5

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. [img]../../data/images/1016-1.gif[/img]

 

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

 

思路：

         每次选出一个与前一个之和是素数的数字放进去，第n个放进去后还要判断与第一个之和是素数才能输出。

代码如下：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
int a[25];
int n ;
int vist[25];
int cases = 1;
void init()
{
   for(int i = 0; i < 25; i ++)
        vist[i] = 0;
    a[0] = 1;
    vist[1] = 1;
}

int is_prime(int x)
{
    for(int i = 2; i <= floor(sqrt(x)+0.5); i ++)
        if(x % i == 0) return 0;
    return 1;
}
void dp(int cur)
{
    if(cur == n && is_prime(1+a[n-1]))
    {
        for(int i = 0; i < n; i ++){
            if(!i)printf("%d",a[i]);
            else
                printf(" %d",a[i]);
        }
        printf("\n");
    }
    else
    {
        for(int i = 1; i <= n; i ++)
        {
            if(!vist[i] && is_prime(i+a[cur-1])){
                a[cur] = i;
                vist[i] = 1;
                dp(cur+1);
                vist[i] = 0;
            }
        }
    }
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        printf("Case %d:\n",cases++);
        init();
        dp(1);
        printf("\n");
    }
    return 0;
}