HDU 1016 Prime Ring Problem (DFS)

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31031    Accepted Submission(s): 13755

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

 

 

 

 

简单DFS，给新生讲课用，题目要求顺时针和逆时针输出，实际好像不需要考虑这一点，最后的情况刚好是这样，素数判断用平方根内有无能整除此数的数字来判断。

#include <iostream>
#include <cmath>
using    namespace    std;

int    N;
int    ANS[25];
bool    VIS[25];

bool    isprimer(int n);
void    dfs(int len);
int    main(void)
{
    int    count = 0;

    while(cin >> N)
    {
        count ++;
        fill(VIS,VIS + 22,false);
        cout << "Case " << count << ":" << endl;
        dfs(0);
        cout << endl;
    }

    return    0;
}

void    dfs(int len)
{
    if(len == N && isprimer(ANS[len - 1] + ANS[0]))
    {
        for(int i = 0;i < N - 1;i ++)
            cout << ANS[i] << ' ';
        cout << ANS[N - 1];
        cout << endl;
        return    ;
    }

    for(int i = 1;i <= N;i ++)
    {
        if(!len && i > 1)
            return;
        if(len && !VIS[i])
        {
            if(isprimer(i + ANS[len - 1]))
            {
                ANS[len] = i;
                VIS[i] = true;
                dfs(len + 1);
                VIS[i] = false;
            }
        }
        else    if(!len)
        {
            ANS[len] = i;
            VIS[i] = true;
            dfs(len + 1);
            VIS[i] = false;
        }
    }

}

bool    isprimer(int n)
{
    for(int i = 2;i <= sqrt(n);i ++)
        if(n % i == 0)
            return    false;
    return    true;
}