HDU 1016 素数环（深搜）

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25134    Accepted Submission(s): 11222


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

 1 #include <stdio.h>
 2 #include <iostream>
 3 #include <string.h>
 4 #include <algorithm>
 5 #include <math.h>
 6 #include <vector>
 7 using namespace std;
 8 
 9 int used[20];
10 int p[40];
11 int n;
12 vector<int>a;
13 int kase;
14 void dfs(int k)
15 {
16     if(k>=n)
17     {
18         if(!p[a[0]+a[n-1]])                 //数组首位相连构成环，首位相加为素数才满足条件 
19         {
20             printf("%d",a[0]);
21             for(int i=1;i<a.size();i++)
22         printf(" %d",a[i]);
23         printf("\n");
24         }
25         return;
26     }
27     for(int i=2;i<=n;i++)
28     {
29         int b=a[a.size()-1];
30             if(!used[i]&&!p[b+i])
31             {
32                 a.push_back(i);
33                 used[i]=1;
34                 dfs(k+1);
35                 a.pop_back();
36                 used[i]=0;
37             }
38     }
39 }
40 main()
41 {
42     int i, j, k;
43     memset(p,0,sizeof(p));                            
44     p[1]=1;p[2]=0;
45     for(i=2;i<=40;i++)                     //筛选法求素数 
46     {
47         for(j=2;j*i<=40;j++)
48         p[j*i]=1;
49     }
50     kase=1;
51     while(scanf("%d",&n)==1)
52     {
53         
54             memset(used,0,sizeof(used));
55             used[1]=1;
56             a.clear();
57             a.push_back(1);
58             printf("Case %d:\n",kase++);
59             dfs(1);
60             cout<<endl;
61     }
62 }