HDU 1016 Prime Ring Problem-优快云博客

本文介绍了一道名为PrimeRingProblem的编程题目，该题目要求在N个数的全排列中找出符合条件的素数组。通过使用深度优先搜索算法（DFS），在排列过程中检查每个可能的组合是否满足相邻两数之和为素数的要求。

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Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34729 Accepted Submission(s): 15381

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

Source

Asia 1996, Shanghai (Mainland China)

Recommend

JGShining

　相对简单的一道题，dfs深搜。即在N个数的全排列，找出符合提议的素数组。

注意：一边排列一边检查

#include<stdio.h>
#include<math.h>
int num;
bool use[20];
int CP[20];
bool isprime(int a)
{
	for(int i=2;i<=sqrt(a+0.0);i++)
	{
		if(a%i==0)
			return false;
	}
	return true;

}
void dfs(int n)
{

	if(n==num&&isprime(1+CP[n-1]))
	{
		for(int i=0;i<num;i++)
		{
			printf(i==num-1?"%d\n":"%d ",CP[i]);
		}
	}
	else
	{
		for(int i=2;i<=num;i++)
		{
			if(!use[i]&&isprime(i+CP[n-1]))
			{
				CP[n]=i;
				use[i]=true;
				dfs(n+1);
				use[i]=false;
			}
		}
	}
}
void init()
{
	for(int i=1;i<=num;i++)
	{
		use[i]=false;
	}
	CP[0]=1;
	use[1]=true;
}
int main()
{
	int time=0;
	while(scanf("%d",&num)!=EOF)
	{
		init();
		printf("Case %d:\n",++time);
		dfs(1);
		puts("");	
	}

}

转载于:https://www.cnblogs.com/WillsCheng/p/DFS.html