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最新推荐文章于 2024-07-20 14:45:00 发布

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本文介绍了一个使用深度优先搜索（DFS）算法解决特定环形填数问题的方法。问题要求在一个由n个圆圈组成的环中填入从1到n的自然数，使得任意相邻两个圆圈内的数字之和为素数。文章提供了完整的C++实现代码，并强调了在判断素数过程中需要注意的细节。

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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

一道简单的DFS递归

检查一个数是否为素数时，可以建立一个数组但是要注意n的范围是<20 所以这个数组的大小最小应该是40 而不是20！我就犯了这个毛病开始WA了半天。

代码如下：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<stdio.h>
#include<cmath>
#include<vector>
#include<list>
using namespace std;
bool Prime[41] = {0,1,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,
                   1,0,0,0,0,0,1,0,0,0};
int Visited[21];
int t[21];
int n;
void DFS(int k)
{
   if(k == n)
   {
       int i;
       for( i = 0;i < n-1;i++)
       {
           cout<<t[i]<<" ";
       }
       cout<<t[n-1]<<endl;
       return;
   }
   for(int j = 2;j <= n;j++)
   {
       if(Visited[j] != 1&&Prime[t[k-1]+j])
       {

               if((k == n-1) && (!Prime[1+j]))
               continue;
               t[k] = j;
               Visited[j] = 1;
               DFS(k+1);
               Visited[j] = 0;
       }
   }
}
int main()
{
   int i,j;
   j = 1;
   while(~scanf("%d",&n))
   {
       memset(Visited,0,sizeof(Visited));
       t[0] = 1;
       cout<<"Case "<<j<<":"<<endl;
       j++;
       DFS(1);
       cout<<endl;
   }
   return 0;
}