素数环（C - 暴力求解）

 

素数环（暴力）（紫书194页）

Description

 

A ring is composed of n (even number) circles as shown in diagram. Put natural numbers 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 <= 16)

n (0 < n <= 16)

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.   

 

You are to write a program that completes above process.

 

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
 
题意：
输入正整数n，把1—n组成一个环，使相邻的两个整数为素数。输出时从整数1开始，逆时针排列。恰好能构成一个环输出一次，n (0 < n <= 16)


分析：
每个环对应于1——n的一个排列，但排列总数高达16！=2*10^13，普通写法会超时，应用回溯法。
DFS。深度优先遍历。

代码：

#include<cstdio>
#include<iostream>
using namespace std;

int n, A[50], isp[50], vis[50];

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



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