Prime is problem - 素数环问题

题目描述：

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.

输入：

n (1 < n < 17).

输出：

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.

样例输入：

6
8

样例输出：

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

---

为了解决该问题，我们可以采用回溯法枚举每一个值。当第一个数位为1确定时，我们尝试放入第二个数，使其和1的和为素数，放入后再尝试放入第三个数，使其与第二个数的和为素数，直到所有的数全部被放入环中，且最后一个数与1的和也是素数，那么这个方案即为答案，输出；若在尝试放数的过程中， 发现当前位置无论放置任何之前未被使用的数均不可能满足条件，那么我们回溯 改变其上一个数，直到产生我们所需要的答案，或者确实不再存在更多的解。

#include "stdafx.h"
#include <stdio.h>
using namespace std;

int prime[] = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37 };
int number[20];
bool hash[20];
int n;
bool isPrime(int num)
{
    for (int i = 0; i < 12;i++)
        if (num == prime[i])
            return true;
    return false;
}

void printArray()
{
    if (isPrime(number[n] + number[1]) == false)
        return;
    for (int i = 1; i <= n; i++)
    {
        if (i != 1)
            printf(" ");
        printf("%d", number[i]);
    }
    printf("\n");
}

void DFS(int num)
{
    if (num > 1 && isPrime(number[num] + number[num - 1]) == false)
        return;
    if (num == n)
    {
        printArray();
        return;
    }

    for (int i = 2; i <= n; i++)
    {
        if (hash[i] == false)
        {
            hash[i] = true;
            number[num + 1] = i;
            DFS(num + 1);
            hash[i] = false;
        }
    }
}

int main()
{
    int cas = 0;
    while (scanf("%d", &n) != EOF)
    {
        cas++;
        for (int i = 0; i < 20; i++)
            hash[i] = false;
        number[1] = 1;
        printf("Case %d:\n", cas);
        hash[1] = true;
        DFS(1);
        printf("\n");
    }

    return 0;
}