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;
}