hdu 1016 Prime Ring Problem dfs

                                           Prime Ring Problem

                                           Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                        Total Submission(s): 36533    Accepted Submission(s): 16102


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 题目意思： 给你一个整数 n 求出所有由 数字1 2 3 4 5 6..... n 中锁有数字组成的素数环并输出出。  素数环的定义： 环中任意相邻的两个数的和是一个素数； 思路： 直接dfs 搜一遍就可以了， n最大是20 不会超时 我的代码：

#include
using namespace std;

const int maxn = 22;
int vis[maxn];
int n;
int ans[maxn];

int cheak(int k)          //函数功能是 检查k是不是素数
{
    int falt = 1;
    for(int i = 2; i < k; i++)
        if(k % i == 0)
           falt = 0;
    return falt;
}

int dfs(int t)
{
    if(t == n +1 && cheak(ans[1] + ans[n]))   //如果检查成功了就输出环
    {
        for(int i = 1 ; i <= n; i++)
             i == n ? printf("%d\n", ans[i]) : printf("%d ", ans[i]);
    }
    for(int i = 2; i <= n; i++)               //位置i的所有可能
    {
        if(cheak(i + ans[t - 1])&& !vis[i])    //检查
        {

            vis[i] = 1;                        //DFS  这里注意设为1， 为了避免重复及死递归
            ans[t] = i;
            dfs(t+1);
            vis[i] = 0;                        //DFS  这里注意归0， 为了避免重复及死递归
        }
    }
}

int main()
{
   int Case = 1;
   while(scanf("%d", &n) == 1 && n)
   {
       memset(vis, 0, sizeof(vis));  //将访问数组归0；
       ans[1] = 1;                   // 1 被固定， 故在开始就设为1
       vis[1] = 1;                   //1 被固定， 故在开始就设为1
       printf("Case %d:\n", Case++); 
       dfs(2);                       // 从2 开始扫, 因为1被固定了
       printf("\n");
   }
}