HDU 1016 素数环(DFS)
Prime Ring Problem
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.
![](http://acm.hdu.edu.cn/data/images/1016-1.gif)
Note: the number of first circle should always be 1.
![](http://acm.hdu.edu.cn/data/images/1016-1.gif)
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
如果是奇数直接No Answer,搜索的时候加上判断当前项和前一项和是否是素数的剪枝。
1 #include<cstdio> 2 #include<cstring> 3 using namespace std; 4 5 int n,a[21],p[41]={0}; 6 bool used[21]; 7 bool flag,key; 8 9 void init() 10 { 11 for(int i=2;i*i<40;i++) 12 for(int j=i+i;j<40;j+=i) 13 p[j]=1; 14 } 15 16 bool check(void) 17 { 18 for(int i=0;i<n;i++) 19 if(p[a[i]+a[(i+1)%n]]) 20 return false; 21 return true; 22 } 23 24 void dfs(int cur) 25 { 26 int i; 27 if(cur==n) 28 { 29 if(check()) 30 { 31 flag=true; 32 for(i=0;i<n-1;i++) 33 printf("%d ",a[i]); 34 printf("%d\n",a[i]); 35 } 36 return; 37 } 38 for(i=2;i<=n;i++) 39 { 40 if(!used[i]&&!p[i+a[cur-1]]) 41 { 42 a[cur]=i; 43 used[i]=true; 44 dfs(cur+1); 45 used[i]=false; 46 } 47 } 48 } 49 50 int main() 51 { 52 int i,j; 53 init(); 54 int kase=0; 55 while(scanf("%d",&n)!=EOF) 56 { 57 a[0]=1,flag=false,used[1]=true; 58 memset(used,0,sizeof(used)); 59 if(n%2&&n!=1) 60 { 61 printf("No Answer\n\n"); 62 continue; 63 } 64 printf("Case %d:\n",++kase); 65 dfs(1); 66 if(!flag) 67 printf("No Answer\n"); 68 printf("\n"); 69 } 70 return 0; 71 }