HDOJ-1016-Prime Ring Problem（素数环）【DFS】

最新推荐文章于 2020-01-29 16:54:27 发布

valar丶morghulis

最新推荐文章于 2020-01-29 16:54:27 发布

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分类专栏： Graph theory

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本文介绍了如何使用深度优先搜索（DFS）算法解决HDOJ1016问题，该问题涉及将连续整数放入圆圈中，使得相邻圆圈的数字之和为质数。通过创建质数筛选函数、递归实现DFS搜索过程，并确保输出符合特定格式，实现了问题的高效求解。

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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.

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

/* Function Description:   HDOJ 1016
   Environment:        DEV C++ 5.6.3.1  
   Technique:          ALGORITHM OF DFS
   Version:  			WATER
   Author:                 valar morghulis  
   Date:                   20150806  
   Notes: 关于这道DFS经典题目，需要注意的是
   1.ONLY存在偶数情况、
   2.输出的格式问题，一不留神就PE 
思路为：筛选出N以内的ALL PRIME，然后从1开始，深搜，此题不用剪枝，很好AC 
*/


#include<cstdio>
#include<cmath>
int a[22];
int use[22];
int n;
int  isprime(int num)//chuangjian yi ge function choice all the prime 
{
    int i;
    for(i=2;i<=sqrt(num+0.0);i++)
    {
        if(num%i==0)
            return 0;
    }
    return 1;//dao ci wei zhi sign wei 1 de wei prime else bushi prime
}
void dfs(int num)
{
    int i;
    if(n==num&&isprime(a[0]+a[n-1]))//ci chu judge wheather first && last number de he wei prime
    {
        for(i=0;i<n;i++)
        {    
       
			if(i==n-1)
			{
				printf("%d\n",a[i]);//form  is very important !!  BE CAREFUL!!
			}
			else
			{
				printf("%d ",a[i]);
			}
 		//	printf(i==n-1?"%d\n":"%d ",a[i]);/THIS IS ANOTHER PRINT WAY ,YOU CAN CHOICE IT !
        }
       // printf("\n");
    }
    else
    {
        for(i=2;i<=n;i++)
        {
            if(!use[i]&&isprime(i+a[num-1]))
            {
                a[num]=i;
                use[i]=1;
                dfs(num+1);//dfs next number 
                use[i]=0;// don't forget huisu huan yuan it 
            }
        }
    }
}

int main()
{
    int cnt = 1; 
     while(~scanf("%d",&n))
    {    
        int i;
        for(i=0;i<n;i++)
        use[i]=0;
        a[0]=1;
        printf("Case %d:\n",cnt++);
        dfs(1);
      //  puts("  ");
      printf("\n");
    }
    return 0;
}