UVa 524 Prime Ring Problem (数论&DFS)-优快云博客

本文介绍了一种解决Prime Ring Problem的算法，通过回溯法寻找满足条件的环形排列，利用素数判断和递归实现解决方案。适用于数学算法和计算机科学领域的专业人士。

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524 - Prime Ring Problem

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=465

A ring is composed of n (even number) circles as shown in diagram. Put natural numbers $1, 2, \dots, n$ 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.

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

完整代码：

/*0.268s*/

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

int n, A[50], isp[50], vis[50];

int is_prime(int x)
{
	for (int i = 2; i * i <= x; i++)
		if (x % i == 0)
			return 0;
	return 1;
}

void dfs(int cur)
{
	if (cur == n && isp[A[0] + A[n - 1]])///递归边界，别忘了测试第一个数和最后一个数
	{
		printf("%d", A[0]);
		for (int i = 1; i < n; i++)
			printf(" %d", A[i]);///打印方案
		putchar('\n');
	}
	else
	{
		for (int i = 2; i <= n; i++)///尝试放置每个数i
			if (!vis[i] && isp[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] = 1;///使用此数并传入后续计算
				dfs(cur + 1);
				vis[i] = 0;///撤销此数的使用
			}
	}
}

int main(void)
{
	int k = 0;
	int first = 0;
	while (~scanf("%d", &n))
	{
		if (first)
			putchar('\n');
		printf("Case %d:\n", ++k);
		memset(A, 0, sizeof(A));
		memset(isp, 0, sizeof(isp));
		for (int i = 2; i <= n * 2; i++)
			isp[i] = is_prime(i);
		memset(vis, 0, sizeof(vis));
		A[0] = 1;
		dfs(1);
		first = 1;
	}
	return 0;
}