HDOJ 1016 Prime Ring Problem素数环【深搜】

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

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        int Case = 0;
        Scanner sc = new Scanner(System.in);
        while(sc.hasNext()){
            int n = sc.nextInt();
            int a[] = new int[n];
            //初始数组1-n
            int color[] = new int[n];
            //判断数字是否已经存在
            int prant[] = new int[n];
            //输出数据排序
            int count =0;//计数器
            for(int i=0;i<n;i++){
                a[i]=i+1;
                color[i] = -1;
            }//初始化数据

            Case++;
            System.out.println("Case "+(Case)+":");

            dfs(a,color,prant,count,0);
            System.out.println();



        }
    }

    private static void dfs(int[] a, int[] color, int[] prant, int count,int m) {
        //System.out.println(count);
        count++;//计数器加1
        if(count == a.length&&p(prant[0],a[m])){
        //注意第一个数和最后一个数相加的和也必须为素数
            prant[count-1]=a[m];
            for(int i=0;i<a.length-1;i++){
                System.out.print(prant[i]+" ");
            }
            System.out.println(prant[a.length-1]);

            //return ;
        }

        for(int i=0;i<a.length;i++){
            color[m] =1;
            if(p(a[m],a[i])&&color[i]==-1){
                color[i]=1;
                prant[count-1]=a[m];

                dfs(a,color,prant,count,i);

                color[i]=-1;


            }


        }


    }

//判断是不是素数
    private static boolean p(int i, int j) {
        int sum = i+j;
        for(int a=2;a*a<=sum;a++){
            if(sum%a==0){
                return false;
            }
        }
        return true;
    }


}

C语言：

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;

int n;
int df[21];
int t=1;
int m[21];
int mi;
bool pn(int x,int y){//判断素数
    for(int i=2;i*i<=x+y;i++){
        if((x+y)%i==0){
            return false;
        }
    }
    return true;
}

void dfs(int x){
    if(mi==n&&pn(m[1],m[n])){
        for(int i=1;i<n;i++){
            printf("%d ",m[i]);
        }
        printf("%d
",m[n]);
        return;
    }


    for(int i=2;i<=n;i++){
        if(df[i]==0&&pn(x,i)){
            df[x]=1;
            mi++;//当前小球数
            m[mi]=i;
            dfs(i);
            df[x]=0;
            mi--;//必须减一
        }
    }


}

int main()
{
    while(~scanf("%d",&n)){
        printf("Case %d:
",t);
        t++;
        memset(df,0,sizeof(df));
        mi=1;
        m[mi]=1;
        dfs(1);
        printf("
");
    }
    return 0;
}