HDU1016 Prime Ring Problem DFS

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 70254 Accepted Submission(s): 30028

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

问题连接

问题描述

输入一个整数n，然后把整数1至n按环放置，要求环上任意两相邻数之和要为素数。如果有多种情况则按字典编撰的顺序输出。

问题分析

第一次见这种题，看网上的题解竟然也是用搜索。因为DFS写的比较少，把程序写完了还楞了一下…
由于n最大为20，所以两数可能的最大值为20+19==39.
用个isprime[40]数组和prime()函数来判断并记录2至39内的素数。
用a[21]记录搜索过程中每一步选取的数。
用vis[21]记录使用过的数。
注意回溯还有输出格式(这里试了两次才通过…)。不管还有没有下一组数据，即跳出循环，都要输出换行。

c++程序如下

#include<iostream>
#include<cmath>
#include<cstdio>
using namespace std;
int n,a[21];
bool isprime[40]={0},vis[21]={0};
void dfs(int);
void prime();
int main()
{
	int t=1;
	prime();
	a[1]=1;
	vis[1]=1;
	while(scanf("%d",&n)!=EOF)
	{
		printf("Case %d:\n",t);
		t++;
		dfs(1);
		printf("\n");
	}
}
void prime()//制素数表
{
	int i,j,k;
	bool flag;
	for(i=2;i<=39;i++)
	{
		k=sqrt((double)i);
		flag=1;
		for(j=2;j<=k;j++)
		{
			if(i%j==0)
			{
				flag=0;
				break;
			}
		}
		if(flag) isprime[i]=1;
	}
}
void dfs(int step)
{
	if(step==n&&isprime[a[n]+a[1]])
	{
		int i,j;
		for(i=1;i<n;i++)
		printf("%d ",a[i]);
		printf("%d\n",a[n]);
	}
	else
	{
		int i,j;
		for(i=2;i<=n;i++)
		{
			if(!vis[i]&&isprime[a[step]+i])
			{
				vis[i]=1;
				a[step+1]=i;
				dfs(step+1);
				vis[i]=0;//回溯
			}
		}
	}
}