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:
 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<algorithm>
 5 using namespace std;
 6 int prime[]={2,3,5,7,11,13,17,19,23,29,31,37};
 7 bool isprime[40];
 8 bool used[21];
 9 int num[21],n;
10 bool dfs(int count,int cur){ //深度优先搜索
11     num[count]=cur;
12     if(count==n-1){
13         if(!isprime[cur+1])return false;
14         for(int i=0;i<n-1;i++)
15             printf("%d ",num[i]);
16         printf("%d\n",num[n-1]);
17         return false;
18     }
19     for(int i=2;i<=n;i++){
20         if(used[i])continue;
21         used[i]=true;
22         if(isprime[cur+i]&&dfs(count+1,i))
23             return true;
24         used[i]=false;
25     }
26     return false;
27 }
28 int main(){
29     memset(isprime,false,sizeof(isprime));
30     for(int i=0;i<12;i++)
31         isprime[prime[i]]=true;//存在素数isprime数组标记为1
32     int Case=1;
33     while(~scanf("%d",&n)){
34         printf("Case %d:\n",Case++);
35         memset(used,false,sizeof(used));
36         used[1]=true;
37         dfs(0,1);
38         printf("\n");
39     }
40     return 0;
41 }

 

 

 

 

posted @ 2017-03-01 18:25  ZERO-  阅读(218)  评论(0)  编辑  收藏  举报