HDOJ1016 Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12758    Accepted Submission(s): 5785

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

 

// 1：当n为奇数的时候， 肯定没有符合条件的:2：符合条件的排列相邻的数字的奇偶性肯定不同；

#include<stdio.h>
#include<string.h>

int used[20],prime[40],path[20];
void DFS(int num,int n)
{
    int i;
    if(num>n&&prime[path[n]+path[1]])//判断第一个和最后一个数是否匹配
    {
        for(i=0;i<n;++i)
            printf(i?" %d":"%d",path[i+1]);
        printf("\n");
        return;
    }
    for(i=2;i<=n;++i)//生成开头数字为1的所有排列
    {
        if(!used[i]&&prime[path[num-1]+i])//如果相邻两个数的和为素数,继续 
        {
            used[i]=1;//将i标记为已用；
            path[num]=i;
            DFS(num+1,n);
            used[i]=0;//恢复原始状态
        }
    }
}

int main()
{
    int n,num=1;
    prime[2]=prime[3]=prime[5]=prime[7]=prime[11]=prime[13]=prime[17]=prime[19]=
    prime[23]=prime[29]=prime[31]=prime[37]=1;
    path[1]=1;
    while(scanf("%d",&n)!=EOF)
    {
        printf("Case %d:\n",num++);
        if(n%2==0)//当n为奇数时，肯定不符合条件；
        {
            memset(used,0,sizeof(used));
            DFS(2,n);//从第二个数开始搜索
        }
        printf("\n");
    }
}