HDU1016

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43299    Accepted Submission(s): 19214

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

素数环问题，相邻两数的和是素数。
素数打表，dfs经典问题
注意dfs的步骤，标记，深搜，消除标记

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

int mark[25], num[25];
int prime[12]={2,3,5,7,11,13,17,19,23,29,31,37};
int n;

bool is_prime(int x)
{
    int i;
    for(i=0;i<12;i++)
        if(x==prime[i]) return true;
    return false;
}

void display()
{
    int i;
    for(i=1;i<n;i++)
        printf("%d ",num[i]);
    printf("%d\n",num[n]);
}

int dfs(int k)
{
    int i;
    if(k==n+1)
    {
        if(is_prime(num[k-1]+1)) display();
    }
    else
    {
        for(i=2;i<=n;i++)
        {
            if(mark[i]&&is_prime(i+num[k-1]))
            {
                mark[i]=0;
                num[k]=i;
                dfs(k+1);
                mark[i]=1;
            }
        }
    }
    return 0;
}

int main()
{
    //if(is_prime(37)) cout<<"*****"<<endl;
    int cnt=1;
    int i;
    while(scanf("%d",&n)!=EOF)
    {
        //cout<<n<<endl;
        printf("Case %d:\n",cnt++);
        if(n==1) {printf("1\n\n");continue;}
        mark[1]=0;
        for(i=2;i<=n;i++)
            mark[i]=1;
        num[1]=1;
        //for(i=1;i<=n;i++) cout<<mark[i];
        //for(i=1;i<=n;i++) cout<<num[i];
        dfs(2);
        printf("\n");
    }
    return 0;
}