HDU 1016.Prime Ring Problem-素数环，相邻两数和为素数-DFS

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 67252    Accepted Submission(s): 28829

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

 

Source

Asia 1996, Shanghai (Mainland China)

 

 

 

 

 

代码:

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 int prime[]={,,,,,,,,,,,};
 bool isprime[];
 bool used[];
 int num[],n;
 bool dfs(int count,int cur){ //深度优先搜索
     num[count]=cur;
     if(count==n-){
         if(!isprime[cur+])return false;
         for(int i=;i<n-;i++)
             printf("%d ",num[i]);
         printf("%d\n",num[n-]);
         return false;
     }
     for(int i=;i<=n;i++){
         if(used[i])continue;
         used[i]=true;
         if(isprime[cur+i]&&dfs(count+,i))
             return true;
         used[i]=false;
     }
     return false;
 }
 int main(){
     memset(isprime,false,sizeof(isprime));
     for(int i=;i<;i++)
         isprime[prime[i]]=true;//存在素数isprime数组标记为1
     int Case=;
     while(~scanf("%d",&n)){
         printf("Case %d:\n",Case++);
         memset(used,false,sizeof(used));
         used[]=true;
         dfs(,);
         printf("\n");
     }
     return ;
 }