hdu 1016 Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 53931    Accepted Submission(s): 23862

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

 

 

直接dfs就可以了

 

 

 

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <cstring>
#include <cmath>
using namespace std;
int vis[25],n;
bool prime(int k)
{
    for(int i=2;i<=sqrt(k);i++)
    {
        if(k%i==0)
        return false;
    }
    return true;
}
void dfs(int k)
{
    if(vis[k]==n&&prime(k+1))
    {
        printf("1");
        for(int i=2;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(vis[j]==i)
                printf(" %d",j);
            }
        }
        printf("
");
        return ;
    }
    for(int i=1;i<=n;i++)
    {
        if(prime(i+k)&&!vis[i])
        {
            vis[i]=vis[k]+1;
            dfs(i);
            vis[i]=0;
        }
    }
}
int main() {
    int Case=0;
    while(~scanf("%d",&n))
    {
        printf("Case %d:
",++Case);
        memset(vis,0,sizeof(vis));
        vis[1]=1;
        dfs(1);
        printf("
");
    }
    return 0;
}