hdu 1016 Prime Ring Problem

[align=left]Problem Description[/align]
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.



[align=left]Input[/align]
n (0 < n < 20).

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

6
8

[align=left]Sample Output[/align]

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

思路：终于考完了冬季“大雪崩”的考试，现在水平也就做做基础题了，本题dfs，用筛法筛出质数然后每次dfs当前的没有被用过数和层数，当转一圈层数正好等于n时再输出，用fa[]记录每个点的父亲结点方便输出，不符合条件的dfs之后要消去访问标记。

#include<iostream>
#include<cstring>
#define CLR(a,b) memset(a,b,sizeof(a))
using namespace std;
const int MAXN=2e2+5;
int n,phi[MAXN],ans[MAXN],fa[MAXN],vis[MAXN];
void eular(){
for(int i = 0; i<MAXN;i++) phi[i] = i;
for(int i = 2; i<MAXN;i++)
if(phi[i] == i)
for(int j = i+i;j<MAXN;j+=i)
phi[j] = 0;
}
void print(int cur){
int n=0;
while(cur!=1){
ans[n++]=cur;
cur=fa[cur];
}
cout<<1;
for(int i=n-1;i>=0;i--)
cout<<' '<<ans[i];
cout<<endl;
}
void dfs(int cur,int k){
if(k==n && phi[cur+1]){
print(cur);
}
//cout<<cur<<' '<<k<<endl;
for(int i=2;i<=n;i++){
if(phi[cur+i] && !vis[i]){
vis[i]=1;
fa[i]=cur;
dfs(i,k+1);
vis[i]=0;
}
}
}

void init(){
CLR(ans,0);CLR(fa,0);CLR(vis,0);
}
int main(){
int ca=1;
eular();
while(cin>>n)
{
cout<<"Case "<<ca++<<":"<<endl;
init();
dfs(1,1);
cout<<endl;
}
return 0;
}