HDU 1016 && Uva 524 Prime Ring Problem(素数环)

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. 


 
Inputn (0 < n < 20). 
OutputThe 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

题目大意 && 解题思路：输入正整数n，把1，2，3，4. . .n组成一个环，使得相邻两个整数之和均为素数。输出从整数1开始的逆时针序列。如果一一枚举，则运算量为n！，并逐个测试会超时。因此，利用回溯法，尝试在cur位置填入可行的数字，再递归调用，在cur+1位置填入可行的数字，直至cur==n输出可行解，注意递归调用后要将vis值恢复。
上代码：#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
using namespace std;

const int maxn=21;
int A[maxn];
int vis[maxn];
int n;
int isp(int x){
if(x==1) return 0;
for(int
4000
i=2;i<=sqrt(x);i++){
if(x%i==0) return 0;
}
return 1;
}
void dfs(int cur){
if(cur==n && isp(A[0]+A[n-1])){
for(int i=0;i<n;i++) {printf("%s%d",!i?"":" ",A[i]);}
printf("\n");
}
else for(int i=2;i<=n;i++)
if(!vis[i] && isp(i+A[cur-1])){
A[cur]=i;
vis[i]=1;
dfs(cur+1);
vis[i]=0;
}
}

int main(){
int kase=0;
while(scanf("%d",&n)!=EOF){
if(kase) printf("\n");
printf("Case %d:\n",++kase);
memset(A, 0, sizeof(A));
memset(vis, 0, sizeof(vis));
A[0]=1;
dfs(1);

}
}