HDU 1016 Prime Ring Problem (DFS(深度優先搜尋))

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20489 Accepted Submission(s): 9159

Problem DescriptionA 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.

Inputn (0 < n < 20).

OutputThe 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

68

Sample Output

Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2

 

題目大意： 判斷 1-n   相鄰兩個數相加的和是不是素數，輸出素數環
DFS 演算法

import java.io.*;import java.util.*;public class Main {public static int n;public static boolean mark[];public static int a[];public static PrintWriter pw;public static void main(String[] args) {Scanner sc=new Scanner(new BufferedInputStream(System.in));//輸入pw=new PrintWriter(new BufferedOutputStream(System.out),true);//輸出int number=1;while(sc.hasNextInt()){n=sc.nextInt();mark=new boolean[n+1];a=new int[n+1];mark[1]=true;a[0]=1;pw.println("Case "+number+++":");DFS(1);pw.println();}}public static void DFS(int len){if(len==n){//判斷第一個元素和最後一個元素是不是素數if(isPrime(a[0]+a[n-1])){for(int i=0;i<n-1;i++){pw.print(a[i]+" ");}pw.println(a[n-1]);}return;}for(int i=2;i<=n;i++){if(!mark[i]&&isPrime(a[len-1]+i)){//判斷相鄰兩個元素是不是素數mark[i]=true;a[len]=i;DFS(len+1);mark[i]=false;//回溯}}}//判斷是不是素數public static boolean isPrime(int m){for(int i=2;i<=Math.sqrt(m);i++){if(m%i==0)return false;}return true;}}