Prime Ring Problem(dfs水)

题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=1016

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34554    Accepted Submission(s): 15303

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

 

Source

Asia 1996, Shanghai (Mainland China)

 简单的dfs

直接上代码:

这里介绍一个报错:Floating point exception (core dumped) linux下报这个一般就是出现了除0或者模0操作....写代码要仔细呀

#include<cstdio>
#include<cstring>
using namespace std;
#define N 50
int a[N];
bool vis[N];
bool is_prime[N];
void init()
{
    for(int i =0 ;i < N ;i++) is_prime[i] = 1;
    for(int i =2  ;i < N ;i++)
    {
        for(int j = 2 ; j <= i/2 ;j++)
        {
            if(i%j==0) {is_prime[i] = 0 ; continue;}
        }
    }
}
void dfs(int n, int cnt)
{
    if(cnt == n&&is_prime[a[0]+a[n-1]]) 
    {
        for(int i =0 ; i < n-1 ;i++)
        {
            printf("%d ",a[i]);
        }
        printf("%d
",a[n-1]);
    }

    for(int i = 1 ;i <= n ;i++)
    {
        if(!vis[i]&&is_prime[a[cnt-1]+i])
        {
            a[cnt] = i;
            vis[i] = 1;
            dfs(n,cnt+1);
            vis[i] = 0;
        }
    }
}

int main()
{
    int n;
    int c = 0;
    init();
    while(~scanf("%d",&n))
    {
        printf("Case %d:
",++c);
        a[0] = 1;
        memset(vis,0,sizeof(vis));
        vis[1] = 1;
        dfs(n,1);
        puts("");
    }
    return 0;
}