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.
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.
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
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 ",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: ",Case++); 35 memset(used,false,sizeof(used)); 36 used[1]=true; 37 dfs(0,1); 38 printf(" "); 39 } 40 return 0; 41 }