Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27775 Accepted Submission(s): 12342
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.
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
/*题解：
第一个自己用DFS解决的搜索问题，一次AC，但感觉对搜索仍然不是很懂。
*/
#include<cstdio>
int a[22],vis[22],n;
int prime[12]={2,3,5,7,11,13,17,19,23,29,31,37};
int is_prime(int x)
{
for(int i=2; i*i<=x; i++)
if(x%i==0)
return 0;
return 1;
}
void dfs(int pos)
{
if(pos==n&&a[0]==1&&is_prime(a[0]+a[n-1]))
{
printf("%d",a[0]);
for(int i=1; i<n; i++)
printf(" %d",a[i]);
printf("\n");
return;
}
for(int i=1; i<=n; i++)
{
if(vis[i]&&is_prime(a[pos-1]+i))
{
vis[i]=0;
a[pos]=i;
dfs(pos+1);
vis[i]=1;
}
}
}
int main()
{
int k;
k=1;
while(scanf("%d",&n)!=EOF)
{
printf("Case %d:\n",k++);
for(int i=1; i<=n; i++)
vis[i]=1;
a[0]=1;
dfs(0);
printf("\n");
}
return 0;
}