题目回顾(HDU-1016):
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.
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
源码与解析
#include<iostream>
#include<string.h>
using namespace std;
int judge[41]={0,
0,1,1,0,1,0,1,0,0,0,
1,0,1,0,0,0,1,0,1,0,
0,0,1,0,0,0,0,0,1,0,
1,0,0,0,0,0,1,0,0,0
};//素数的打表
int mark[21];//用作标记是否这个数已经使用过
int y[21];//答案数组
int n;
void dfs(int a){
//如果所有数字确定,那么判断首尾相加是否也是素数,
//若是,则满足条件,输出
if(a==n&&judge[y[1]+y[n]]){
for(int i=1;i<n;i++){
cout<<y[i]<<" ";
}
cout<<y[n]<<endl;
return;
}
//深搜,不断找出未被标记且与此数相加为素数的下一个数
for(int i=1;i<=n;i++){
if(mark[i]==0&&judge[y[a]+i]){
y[a+1]=i;
mark[i]=1;
dfs(a+1);
mark[i]=0;
}
}
}
int main(){
int c=1;
while(cin>>n){
cout<<"Case "<<c++<<":"<<endl;
memset(y,0,sizeof(y));
memset(mark,0,sizeof(mark));
y[1]=1;
mark[1]=1;
dfs(1);
cout<<endl;
}
return 0;
}