hdu 1016 Prime Ring Problem（深度优先搜索）

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 12105    Accepted Submission(s): 5497

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

 

 

题意：输入一个 n 找出1~n的组合，使得相邻两个数之和为素数；

分析：预处理40之间的素数，然后回溯；

#include<iostream>
#include<cstring>
#define N 25
#define M 40
using namespace std;

bool is_prime[M],visited[N];
int n,test,ans[N];

void work(int k)
{
    int i;
    if(k==n+1)
    {
        if(!is_prime[ans[n]+ans[1]]) return ;
        for(i=1;i<=n-1;i++)
            cout<<ans[i]<<" ";
        cout<<ans[i]<<endl;
        return ;
    }
    for(i=2;i<=n;i++)
    {
        if(!visited[i]&&is_prime[ans[k-1]+i])
        {
            visited[i]=true;
            ans[k]=i;
            work(k+1);
            visited[i]=false;
        }
    }
}

bool prime(int n)
{
    if(n==1) return false;
    if(n==2||n==3) return true;
    int i;
    for(i=2;i<n;i++)
        if(n%i==0)
            return false;
    return true;
}

int main()
{
    int i;test=1;
    for(i=1;i<M;i++) is_prime[i]=prime(i);
    while(cin>>n)
    {
        ans[1]=1;
        memset(visited,false,sizeof(visited));
        cout<<"Case "<<test<<":"<<endl;
        work(2);
        test++;
        cout<<endl;
    }
    return 0;
}