Prime Ring Problem(DFS）

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

 68 

 

Sample Output

 Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2 

 

解题思路：

用DFS检索所用的情况，如果满足条件输出即可，因为数不超过20，所以可以枚举出所有40以内的素数。一开始脑子抽筋，只枚举了20以内的素数，WA了一发，没有考虑到虽然数字不超过20，但两个数相加是不超过40，所以得枚举40以内的素数。。。。

AC代码：

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn = 25;int n, num[maxn] = {0,1},prime[11] = {3,5,7,11,13,17,19,23,29,31,37};bool visited[maxn];bool isPrime(int a){    for(int i = 0; i < 11; i++)    {        if(a == prime[i])            return true;    }    return false;}void Dfs(int x){    if(x == n && isPrime(num[x] + 1))  // 最后一个数要判断和1相加是否为素数，这样就构成了环    {        for(int i = 1; i <= n; i++)        {            if(i == n)                printf("%d\n",num[i]);            else                printf("%d ",num[i]);        }        return;    }    for(int i = 2; i <= n; i++)    {        if(!visited[i] && isPrime(num[x] + i))        {            visited[i] = 1;            num[x + 1] = i;            Dfs(x + 1);            visited[i] = 0;        }    }    return;}int main(){    int count = 0;    while(scanf("%d", &n) != EOF)    {        memset(visited, 0, sizeof(visited));        printf("Case %d:\n",++count);        Dfs(1);        printf("\n");    }    return 0;}