Prime Ring Problem 分类： dfs 2015-04-27 17:00 17人阅读 评论(0) 收藏

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

Source
Asia 1996, Shanghai (Mainland China)
Difficulty:dfs 素数的判定；
DFS:初始状态：放了n个数
扩展方式：选一个没有选过的数且与上一个数的和是素数
目标状态：选了n个数，且第n个数和第一个数和是素数（因为首位相连）

#include <cstdio>
#include <cstring>
#include <algorithm>
#include<cmath>
using namespace std;
int prime(int k)
{
    if(k<=1) return 0;
    int m=floor(sqrt(k)+0.5);
    for(int i=2;i<=m;i++)
    if(k%i==0)return 0;
    return 1;
}//素数的判断；
int n;
int a[100];
int vis[100];
void dfs(int num)
{
    int i;
    if(num==n&&prime(a[num-1]+a[0]))//目标状态：选了n个数，且第n个数和第一个数和是素数（因为首位相连）
    {
        for(int i=0;i<n-1;i++)
        printf("%d ",a[i]);
        printf("%d\n",a[num-1]);
    }
    else
    {
        for(i=2;i<=n;i++)
        {
            if(vis[i]==0&&prime(i+a[num-1]))//扩展方式：选一个没有选过的数且与上一个数的和是素
            {vis[i]=1;
             a[num++]=i;
             dfs(num);
             vis[i]=0;
             num--;
            }
        }
    }
    return;
}
int main()
{int c=1;
  while(~scanf("%d",&n))
  {printf("Case %d:\n",c);
  memset(vis,0,sizeof(vis[0]));
  a[0]=1;
  dfs(1);//初始状态：放了n个数
初始状态：放了n个数；
  c++;
  printf("\n");}
  return 0;
}

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