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): 47896    Accepted Submission(s): 21158

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

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

分析：dfs求解，相邻两个数素数的判定（当前数加上之前一个数进行判定，第一个数和最后一个数进行判定）

AC代码：

#include<cstdio>#include<cstring>#include<cmath>using namespace std;const int maxn=20+1;int vis[maxn];int a[maxn];int n;bool is_prime(int i1,int i2){for(int j=2;j<=sqrt(i1+i2);j++)if((i2+i1)%j==0)return false ;return true;}void dfs(int cur){if(cur>n && is_prime(a[1],a[n])){printf("%d",a[1]);for(int i=2;i<=n;i++)printf(" %d",a[i]);printf("\n");}for(int i=1;i<=n;i++){if(!vis[i]){if(is_prime(a[cur-1],i)){a[cur]=i;vis[i]=1;dfs(cur+1);vis[i]=0;}}}}int main(){int cnt=0;while(scanf("%d",&n)==1){  memset(vis,0,sizeof(vis));  a[1]=1;  vis[1]=1;  printf("Case %d:\n",++cnt);  dfs(2);  printf("\n");  }return 0;}