HDU 1016 Prime Ring Problem（深搜+回溯+剪枝+Eratosthenes素数筛）

• Prime Ring Problem

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
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

__题意：__给出一个正整数n，把1-n连成一个环，使得相邻的两个数和为素数，并且从1开始逆时针围成的环，输出可行的方案，并按字典序输出。0 题目有点小坑，它没有说只能数字之间有空格，害我提交的时候PE了
废话不多说，现在来讲思路，首先要用到Eratosthenes素数筛，求出2-40内的素数，因为20+20=40，两数之和最大也不会超过40。然后就开始深搜dfs，剪枝，回溯，详细参见代码。

#include 
#include 

using namespace std;
int prime[40] = {0}, vis[40] = {0}, pre[25], cnt = 0, n;//pre数组存储素数环，vis数组表示访问过的元素，prime数组为0表示素数
void init()
{
	for(int i = 2; i <= 7; i++)
	{
		if(!prime[i])
			for(int j = i * i; j < 40; j += i) prime[j] = 1;
	}
}
void dfs(int cur)
{
	if(cur == n)
	{
		if(!prime[pre[n - 1] + pre[0]])
		{
			for(int i = 0; i < n - 1; i++) printf("%d ", pre[i]);
			printf("%d\n", pre[n - 1]);
		}
	}
	for(int i = 2; i <= n; i++)
	{
		if(!vis[i] && !prime[pre[cur - 1] + i]) //如果没有访问过的数字并且与前一个数字相加为素数，则继续搜下一个，并且将这个数字标记为已访问
		{
			vis[i] = 1;
			pre[cur] = i;
			dfs(cur + 1);
			vis[i] = 0; //完成本次搜索后取消标记，以免下次搜索时没有访问被误判为访问过的数字
		}
	}
}
int main()
{
	init();
	int cnt = 0;
	while(scanf("%d", &n) != EOF){
		printf("Case %d:\n", ++cnt);
		memset(vis, 0, sizeof(vis));
		pre[0] = 1;
		dfs(1);
		printf("\n");
	}
	return 0;
} 

End