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): 46985    Accepted Submission(s): 20746

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

 
思路：
简单的DFS，做过突然想温习一下，旧题新写

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
using namespace std;
const int MAXN=21;
bool flag[MAXN];
int ans[MAXN];
int total=0;
int t=1;
int N;
bool check(int x, int y)
{
    x+=y;
    int i=2;
    while(i<=sqrt(x)&&x%i)
    {
        i++;
    }
    if(i>sqrt(x))
        return true;
    return false;
}
void print()
{
    printf("%d",ans[1]);
    for(int i=2;i<=N;i++)
    {
        printf(" %d",ans[i]);
    }
    printf("\n");
}
void dfs(int num)
{
    for(int i=2;i<=N;i++)
    {
        if(check(ans[num-1],i)&&!flag[i])
        {
            ans[num]=i;
            flag[i]=true;
            if(num==N)
            {
                if(check(ans[num],ans[1]))
                    print();
            }
            else
            {
                dfs(num+1);
            }
            flag[i]=false;
        }
    }
}
int main()
{
    while(scanf("%d",&N)!=EOF)
    {
        memset(flag,false,sizeof(flag));
        printf("Case %d:\n",t);
        ans[1]=1;
        dfs(2);
        printf("\n");
        t++;
    }
    return 0;
}