HDU 1016Prime Ring Problem【深搜】

#include<stdio.h>
#include<string.h>
int prime[40];
int a[20];
int vis[20];
int n;
int isprime(int n)
{
    int i;
    for(i=2;i*i<=n;i++)
    if(n%i==0)
        return 0;
    return 1;
}
void dfs(int cur)
{
    int i;
    if(cur==n&&prime[a[0]+a[n-1]])     //递归边界，最后一个数和第一个数和是质数
    {
        for(i=0;i<n;i++)
         printf("%d%c",a[i],(i==n-1)?'\n':' ');
    }
    else
    for(i=2;i<=n;i++)                         //尝试放置每个数 i
        if(!vis[i]&&prime[i+a[cur-1]])        //如果 i 没有用过，并且与前一个数的和为质数 
        {
        a[cur]=i;
        vis[i]=1;                                        //标记
        dfs(cur+1);
        vis[i]=0;                                       //清除标记
        }
}
int main()
{
    int i,k=0;
    memset(prime,0,sizeof(prime));
    memset(vis,0,sizeof(vis));
    prime[2]=1;
    for(i=3;i<40;i+=2)
        if(isprime(i))
     prime[i]=1;
    while(scanf("%d",&n)!=EOF)
    {
        k++;
        printf("Case %d:\n",k);
        for(i=0;i<n;i++)
            a[i]=i+1;
        dfs(1);
        putchar('\n');
    }
return 0;
} 

Problem 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

总结： 回溯算法