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HDU 1016 Prime Ring Problem C语言版

HDU题解 Slyar 3105浏览 0评论

文章作者:姜南(Slyar) 文章来源:Slyar Home (www.slyar.com) 转载请注明,谢谢合作。

非常典型非常顺手的深度优先搜索(DFS)题,写起来非常痛快。

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

Slyar:大意是给出N,求所有由1-N组成的首位为1环形序列,要求相邻两数的和为素数。素数判断不说了,因为N小于20,所以可以预先将素数打表保存供后面查询。因为规定了首位为1,所以从1开始搜索所有的可能,用visit[]标记访问过的数字,每次搜到深度等于N的时候就可以打印结果path[]了。没有结果的也就空着输出就好了,格式方面注意每行结果的最后一个数字后面没有空格。

#include 
#include 
#include 
#include 
#define MAX 51

int n;
int path[MAX];
int visit[MAX];
int prime[MAX];

int isPrime(int x)
{
	int i;
	for (i = 2; i <= sqrt(x*1.0); i++)
	{
		if (x % i == 0) return 0;
	}
	return 1;
}

void dfs(int x)
{
	int i;
	if ((x == n) && prime[path[1] + path[n]])
	{
		for (i = 1; i < n; i++)
		{
			printf("%d ", path[i]);
		}
		printf("%d\n", path[n]);
		return;
	}
	else
	{
		for (i = 2; i <= n; i++)
		{
			if (!visit[i] && prime[path[x] + i])
			{
				path[x+1] = i;
				visit[i] = 1;
				dfs(x+1);
				visit[i] = 0;
			}
		}
	}
}

int main()
{
	int i, t;
	
	// 打表 
	for (i = 1; i < MAX; i++)
	{
		if (isPrime(i)) prime[i] = 1;
		else prime[i] = 0;
	}
	
	// 输出 
	t = 0;
	while(scanf("%d", &n) != EOF)
	{
		printf("Case %d:\n", ++t);
		memset(visit, 0, sizeof(visit));
		path[1] = 1;
		visit[1] = 1;
		dfs(1);
		printf("\n");
	}
	//system("pause");
	return 0;
}

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