HDU 1016 Prime Ring Problem(DFS入门)

最新推荐文章于 2019-11-23 11:20:08 发布

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最新推荐文章于 2019-11-23 11:20:08 发布

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分类专栏： DFS算法

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本文介绍了一个名为PrimeRingProblem的问题，该问题要求将1至n的自然数放置于由n个圆组成的环中，使得任意相邻两个圆上的数字之和为素数。文章通过两段代码示例详细展示了如何使用深度优先搜索（DFS）算法来寻找所有可能的解决方案，并对代码进行了注释以便理解。

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Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 54078 Accepted Submission(s): 23943

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

DFS入门：注意结束条件

#include <stdio.h>
#include<string.h>
int n;
int a[100],visit[100];
int prime(int x)
{
	int i;
	for(i=2;i*i<=x;i++)
	if(x%i==0)
	return 0;
	return 1;
}
void dfs(int num)
{
	int i;
	if(num==n&&prime(a[num-1]+1))
	{	
		for(i=0;i<n-1;i++)
		printf("%d ",a[i]);
		printf("%d\n",a[num-1]);
	}
	else
	{
		for(i=2;i<=n;i++)
		{
			if(visit[i]==0)
			{
				if(prime(a[num-1]+i))
				{
					visit[i]=-1;
					a[num++]=i;
					dfs(num);
					visit[i]=0;
					num--;
				}
			}
		}
	}
}
int main(int argc, char *argv[])
{
	int flag=1;
	while(scanf("%d",&n)!=EOF)
	{
		memset(visit,0,sizeof(visit));
		a[0]=1;
		printf("Case %d:\n",flag++);
		dfs(1);
		printf("\n");
	}
	return 0;
}

2.0 版本

#include <stdio.h>
#include<algorithm>
#include<string>
#include<set>
using namespace std;
int n;
int visit[300];
int a[300];
bool prime(int n)
{
	int i;
	for(i=2;i*i<=n;i++)
	{
		if(n%i==0)
		return false;
	}
	return true;
}
void dfs(int num) 
{
	int i,j;
	if(num==n&&prime(a[num-1]+1))
	{
		for(i=0;i<n-1;i++)
		printf("%d ",a[i]);
		printf("%d\n",a[n-1]);
	}
	else
	{
		for(i=2;i<=n;i++)//找到了进行下一层循环 找不到后退 向后递归时visit[i]自动变为 0  
		{
			if(prime(i+a[num-1])&&visit[i]==0)
			{
				visit[i]=1;
				a[num]=i;
				dfs(num+1);
				visit[i]=0;
			}
		}
	}
}
int main(int argc, char *argv[])
{
	int flag=1;
	while(scanf("%d",&n)!=EOF)
	{
		memset(visit,0,sizeof(visit));
		a[0]=1;
		printf("Case %d:\n",flag++);
		dfs(1);
	 	printf("\n");
	}
	return 0;
}