UVa 524 Prime Ring Problem（回溯法）

Posted 2020-08-21

传送门

Description

A ring is composed of n (even number) circles as shown in diagram. Put natural numbers 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 ≤ 16)

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. You are to write a program that completes above process.

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

思路

题意：

输入正整数n，把整数1，2，3，……，n组成一个环，使得相邻两个整数之和均为素数。输出时序列头为1开始的序列，同一个环应恰好输出一次

题解:

#include<bits/stdc++.h>
using namespace std;
const int maxn = 40;
bool is_prime[maxn],vis[maxn];
int n,a[maxn];

void dfs(int cur)
{
	if (cur == n && is_prime[a[0] + a[n-1]])  //递归边界，因为是环，所以还要测试第一个和最后一个 
	{
		printf("%d",a[0]);
		for (int i = 1;i < n;i++)	printf(" %d",a[i]);
		printf("\n"); 
	}
	else
	{
		for (int i = 2;i <= n;i++)    //尝试放置每个数i 
		{
			if (!vis[i] && is_prime[i + a[cur-1]])  //如果i没有用过，并且与前一个数之和为素数 
			{
				a[cur] = i;
				vis[i] = true;                     //设置使用标志 
				dfs(cur+1);
				vis[i] = false;                    //清楚标志 
			}
		}
	}
}

int main()
{
	//提前预处理素数表 
	memset(is_prime,true,sizeof(is_prime));
	is_prime[0] = is_prime[1] = false;
	for (int i = 2;i < maxn;i++)
	{
		if (is_prime[i])
		{
			for (int j = 2 * i;j < maxn;j += i)
			{
				is_prime[j] = false;
			}
		}
	}
	int tcase = 0;
	while (~scanf("%d",&n))
	{
		memset(vis,false,sizeof(vis));
		memset(a,0,sizeof(a));
		if (tcase)	printf("\n");
		printf("Case %d:\n",++tcase);
		a[0] = 1;
		dfs(1);
	}
	return 0;
}