HDU 1016 Prime Ring Problem DFS

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.

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

《此处应有空格》

 

这道题目就是让你全排列一个总数字为n个的素数环。

打了一个素数表然后简单的DFS直接就可以过了。

一开始跟样例输出一样，一直不知道为什么PE了，后来看讨论版，发现其实应该在每一个Case后面加空格，而不是两个Case之间有空格。

代码如下：

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
using namespace std;
int n,k;
int num[13]={0,2,3,5,7,11,13,17,19,23,29,33,37};
int check[21];
bool use[21];
int before;
bool panduan(int a)
{
    for (int i=1;i<=12;++i)
    {
        if (a==num[i])
        return true;
    }
    return false;
}
void search(int x,int nn)
{ 
int t;
    if (nn==n)
    {   
        
        if (panduan(x+1)==true)
        {
        
        for(int i=1;i<n;++i)
        printf("%d ",check[i]);
        printf("%d\n",check[n]);
        }
        return;
    }
    for (int i=2;i<=n;++i)
    {
        if (use[i]==false&&panduan(i+before)==true)
        {
            check[nn+1]=i;
            use[i]=true;
            t=before;
            before=i;
            search(i,nn+1);
            check[nn+1]=0;
            use[i]=false;
            before=t;
        }
    }
}
int main(){
    k=0;
    while(scanf("%d",&n)!=EOF)
    {

        for(int i=1;i<=n;++i)
        {
        check[i]=0;
        use[i]=false;
        }
        use[1]=true;
        check[1]=1;
        int xx=1;
        k++;
        
        printf("Case %d:\n",k);
        if(xx!=n)
       {
        before=1;
        search(1,1);
        }else printf("1");
        printf("\n");
    }
    return 0;
}