HDU - 1016-Prime Ring Problem（DFS）

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探讨了在一个由n个圆组成的环形结构中，如何将自然数1至n分配到每个圆中，使得任意相邻两圆内的数字之和为素数的问题。通过深度优先搜索算法（DFS）实现解决方案的查找与输出。

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

我的见解：不用多说，就是dfs；然后要输出好几列满足条件的数字，搞两个数组，循环中用过的数字就标志，没标志的就可以用，然后另一个数组就用来记录满足条件的一列数字，满足题意则输出，也是用常规的递归来完成

#include<bits/stdc++.h>
using namespace std;

int n, k = 0;
int a[22] = { 0 }, b[22] = { 1 ,1};
int pan(int k)
{
	int j;
	for (j = 2; j <= sqrt(k); j++)
	{
		if (k%j == 0)
		{
			return 0;
		}
	}
	return 1;
}

void dfs(int cur)
{
	if (cur == n && pan(1+b[n - 1])==1)
	{
		cout << b[0];
		int i;
		for (i = 1; i < n; i++)
		{
			cout << ' ' << b[i];
		}
		cout << endl;
	}
	else
	{
		for (int i = 2; i <= n;i++)
		{
			if (a[i]==1 || pan(b[cur - 1]+i)==0)
			{
				continue;
			}
			b[cur] = i;
			a[i] = 1;
			dfs(cur + 1);
			a[i] = 0;
		}
	}
}



int main()
{
	
	while (cin >> n)
	{
		k++;
		cout << "Case " << k << ":" << endl;
		dfs(1);
		cout << endl;
	}
}