hdu 1016 Prime Ring Problem（深度優先搜索），hdu深度優先搜索
Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 12105 Accepted Submission(s): 5497
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
題意：輸入一個 n 找出1~n的組合，使得相鄰兩個數之和為素數；
分析：預處理40之間的素數，然後回溯；
![]()
1 #include<iostream>
2 #include<cstring>
3 #define N 25
4 #define M 40
5 using namespace std;
6
7 bool is_prime[M],visited[N];
8 int n,test,ans[N];
9
10 void work(int k)
11 {
12 int i;
13 if(k==n+1)
14 {
15 if(!is_prime[ans[n]+ans[1]]) return ;
16 for(i=1;i<=n-1;i++)
17 cout<<ans[i]<<" ";
18 cout<<ans[i]<<endl;
19 return ;
20 }
21 for(i=2;i<=n;i++)
22 {
23 if(!visited[i]&&is_prime[ans[k-1]+i])
24 {
25 visited[i]=true;
26 ans[k]=i;
27 work(k+1);
28 visited[i]=false;
29 }
30 }
31 }
32
33 bool prime(int n)
34 {
35 if(n==1) return false;
36 if(n==2||n==3) return true;
37 int i;
38 for(i=2;i<n;i++)
39 if(n%i==0)
40 return false;
41 return true;
42 }
43
44 int main()
45 {
46 int i;test=1;
47 for(i=1;i<M;i++) is_prime[i]=prime(i);
48 while(cin>>n)
49 {
50 ans[1]=1;
51 memset(visited,false,sizeof(visited));
52 cout<<"Case "<<test<<":"<<endl;
53 work(2);
54 test++;
55 cout<<endl;
56 }
57 return 0;
58 }
View Code
