Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20966 Accepted Submission(s): 9386
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
Source
简单DFS直接上代码：
#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<cmath>
using namespace std;
int visi[20];
int n;
int res[20];
bool isprime(int u)
{
if(u==0||u==1) return false;
if(u==2) return true;
if(u%2==0) return false;
for(int i=3;i<=sqrt(double(u));i+=2)
if(u%i==0)
return false;
return true;
}
void dfs(int p,int step)
{
int i;
if(step==n)
{
if(isprime(p+1))
{
cout<<res[1];
for(i=2;i<=n;i++)
cout<<" "<<res[i];
cout<<endl;
}
return;
}
for(i=2;i<=n;i++)
{
if(!visi[i]&&isprime(i+res[step]))
{
visi[i]=1;
res[step+1]=i;
dfs(i,step+1);
visi[i]=0;
}
}
}
int main()
{
int cas=0;
while(~scanf("%d",&n))
{
memset(visi,0,sizeof(visi));
printf("Case %d:\n",++cas);
visi[1]=1;
res[1]=1;
dfs(1,1);
cout<<endl;
}
return 0;
}
//890MS 284K