UVA 524 素数环 【dfs/回溯法】

Description

 

A ring is composed of n (even number) circles as shown in diagram. Put natural numbers 1,2,3，...，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 <= 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


【代码】：

/*素数环*/
#include <bits/stdc++.h>
using namespace std;
int n,prime[1005]={1},v[1005],a[1005];

void get_prime()
{
    int m=sqrt(2*n+0.5);//环-倍增
    memset(prime,1,sizeof(prime));//0为素数
    prime[0]=prime[1]=0;//非素数
    for(int i=2;i<=m;i++)
    {
        if(prime[i])//素数
        {
            for(int j=i+i;j<=2*n;j+=i)
            {
                prime[j]=0;//非素数
            }
        }
    }
}

void dfs(int cur)
{
    if(cur == n && prime[ a[0]+a[n-1] ])
    {
        for(int i=0; i<n; i++)
        {
             printf("%d%c", a[i], i == n - 1 ? '
' : ' ');
        }
       // printf("
");
    }
    else
    {
        for(int i=2; i<=n; i++)
        {
            if( !v[i] && prime[ i + a[cur-1] ] )
            {
                a[cur] = i;
                v[i]=1;
                dfs(cur+1);
                v[i]=0;
            }
        }
    }
}
int main()
{
    scanf("%d",&n);
    //初始化
    memset(v,0,sizeof(v));
    a[0]=1;

    get_prime();
    dfs(1);
}