HDU - 1016 Prime Ring Problem

HDU - 1016 Prime Ring Problem
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

给出一个n，和一个环。在坏上放置n个数，即1-n，要求所有相邻的两个数之和都必须是质数，求满足该要求的所有序列，注意序列都是以1开始的。
DFS下去，觉得有点类似n皇后问题吧。
注意递归回溯之后把vis重置，因为回溯之后该点是可以走的。

AC代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <cstdlib>

using namespace std;
int n , a[20],put[20];
bool isPrime(int num){
    for(int k =2 ; k < num ; ++k){
        if(num % k == 0) return false;
    }
    return true;
}
void dfs(int i){//放置第i个数

    for(int j = 2 ; j <= n ; ++j){
        if(i == n+1){
        if(isPrime(1+put[n])){
            for(int j = 1 ; j < n ; ++j){
                    printf("%d ",put[j]);
            }
            printf("%d\n",put[n]);
            return;
        }
    }
        if(a[j]) continue;
        if(isPrime(j+put[i-1])){
            put[i] = j;
            a[j] = 1;
            dfs(i+1);
            a[j] = 0;
        }   
    }
}
int main(){
    int c = 1;
    put[1]= 1;
    while(~scanf("%d",&n)){
        memset(a,0,sizeof(a));
        printf("Case %d:\n",c++);
        dfs(2);
        printf("\n");
    } 
    return 0;
}