hdu1016(简单深搜)

这是一个简单深搜问题，要求相邻的数之间相加为素数。采用深搜，把满足条件的值放在parent[]中。最后输出parent[].

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.

6
8

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

实现代码：

import java.util.Scanner;

public class Main {
	static int k=0;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while (sc.hasNext()) {
			int n = sc.nextInt();  //输入个数
			int[] a = new int[n];  //用来装（1~n）
			int[] color = new int[n]; //用来做标记
			int[] parent = new int[n]; //用来装结果的数组
			int count = 0;   //用来记录搜索个数
			System.out.println("Case "+(++k)+":");
			// 初始化数据
			for (int i = 0; i < n; i++) {
				a[i] = i + 1;
				color[i] = -1;//把未搜索的标记为-1
				parent[i] = -1;
			}
			dfs(a, color, parent, 0, count);
			System.out.println();
		}
	}

	private static void dfs(int[] a, int[] color, int[] parent, int u, int count) {
		color[u] = 1; //把搜索过的标记为1
		count++;
		//递归跳出的条件：次数到达了给定的的个数即数组a[]的长度；最后一个数和第一个数相加满足是素数
		if (count == a.length && isPrime(a[u] + a[0])) {
			parent[count - 1] = a[u];//把最后一个结果放到parent中
			print(parent);//输出结果
			return;
		}
		for (int v = 0; v < a.length; v++) {
			//满足color值为-1和当前值和下一个值相加为素数进入dfs
			if (color[v] == -1 && isPrime(a[v] + a[u])) {
				parent[count - 1] = a[u];//把满足的值放在结果数组中
				dfs(a, color, parent, v, count);
				color[v] = -1;//还原标记
			}
		}
	}
    //输出结果
	private static void print(int[] parent) {
		for (int i = 0; i < parent.length; i++) {
			if (i < parent.length - 1) {
				System.out.print(parent[i] + " ");
			} else {
				System.out.println(parent[i]);
			}
		}
	}

	//判断是否为素数
	private static boolean isPrime(int num) {
		for (int i = 2; i * i <= num; i++) {
			if (num % i == 0) {
				return false;
			}
		}
		return true;
	}
}

hdu1016(简单深搜)

原文：http://blog.csdn.net/xionghui2013/article/details/47210449