HDOJ1016(标准dfs)

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 42688    Accepted Submission(s): 18919

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

 

#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN=;
int n,buf[MAXN],vis[MAXN];
bool isPrime(int x)
{
    if(x<=)    return false;
    for(int i=;i*i<=x;i++)
    {
        if(x%i==)    return false;
    }
    return true;
}
void dfs(int dep)
{
    if(dep==n)
    {
        for(int i=;i<dep-;i++)
        {
            printf("%d ",buf[i]);
        }
        printf("%d\n",buf[dep-]);
        return ;
    }
    for(int i=;i<=n;i++)
    {
        if(!vis[i]&&isPrime(i+buf[dep-]))
        {
            if(dep==n-&&!isPrime(i+buf[]))
                continue;
            vis[i]=;
            buf[dep]=i;
            dfs(dep+);
            vis[i]=;
        }
    }
}
int main()
{
    int cas=;
    while(scanf("%d",&n)!=EOF)
    {
        memset(vis,,sizeof(vis));
        printf("Case %d:\n",++cas);
        buf[]=;
        dfs();
        printf("\n");
    }
    return ;
}