HDU 1016 素数环(DFS)

　　　　　　　　　　　　　　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.

 

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

 

如果是奇数直接No Answer，搜索的时候加上判断当前项和前一项和是否是素数的剪枝。

 

#include<cstdio>
#include<cstring>
using namespace std;

int n,a[21],p[41]={0};
bool used[21];
bool flag,key;

void init()
{
    for(int i=2;i*i<40;i++)
    for(int j=i+i;j<40;j+=i)
    p[j]=1;
}

bool check(void)
{
    for(int i=0;i<n;i++)
    if(p[a[i]+a[(i+1)%n]])
    return false;
    return true;
}

void dfs(int cur)
{
    int i;
    if(cur==n)
    {
        if(check())
        {
            flag=true;
            for(i=0;i<n-1;i++)
            printf("%d ",a[i]);
            printf("%d
",a[i]);
        }
        return;
    }
    for(i=2;i<=n;i++)
    {
        if(!used[i]&&!p[i+a[cur-1]])
        {
            a[cur]=i;
            used[i]=true;
            dfs(cur+1);
            used[i]=false;
        }
    }
}

int main()
{
    int i,j;
    init();
    int kase=0;
    while(scanf("%d",&n)!=EOF)
    {
        a[0]=1,flag=false,used[1]=true;
        memset(used,0,sizeof(used));
        if(n%2&&n!=1)
        {
            printf("No Answer

");
            continue;
        }
        printf("Case %d:
",++kase);
        dfs(1);
        if(!flag)
        printf("No Answer
");
        printf("
");
    }
    return 0;
}