prime ring problem回溯法-优快云博客

本文介绍了一个环形填数问题的解决方法，要求在环形结构中填充从1到n的自然数，相邻两数之和为质数，且首个数字必须为1。通过递归算法和质数判断实现所有可能解的搜索与输出。

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这道题跟八皇后简直一样

可以这么想象，你要填充那个数组，实际上就是下楼梯。

关键是选哪一个楼梯下去。一层有多个楼梯选择。

循环，如果这个楼梯可以下去，那么记录，并下去。

如果到了底部，那么输出。

归零，装作没有来过。

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

#include <iostream>
using namespace std;
int a[18],num_used[18]={0},count=1;
bool is_prime(int n)
{
    for(int i=2;i<=n/2;i++)
    {
        if(n%i==0)return false;
    }
    return true;
}
bool ok(int j,int previous)
{
    bool used=true;
    for(int i=0;i<count;i++)
    {
        if(num_used[i]==j){used=false;break;}
    }

    return used;

}
void find_ring(int n,int height)
{
    for(int j=2;j<=height;j++)
    {
        if(ok((j),a[n-1])&&is_prime(j+a[n-1]))
        {
            a[n]=j;
            num_used[count++]=j;
            find_ring(n+1,height);
        }
    }
    if(n==height)
    {
        if(is_prime(a[n-1]+1)){
        for(int i=0;i<n;i++)cout<<a[i]<<" ";
        cout<<endl;}
    }
    a[n]=0;
    num_used[count-1]=0;
    count--;
}
int main()
{
    int height,ha=1;
    while(cin>>height){
    num_used[0]=1;
    a[0]=1;
    cout<<"Case "<<ha++<<":"<<endl;
    find_ring(1,height);
}
}