HDU - 1016 Prime Ring Problem DFS

最新推荐文章于 2021-02-26 15:30:16 发布

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本文介绍了一种解决PrimeRingProblem的方法，该问题要求在由n个圈组成的环中填入从1到n的自然数，使得相邻两圈的数字之和为素数。通过列出39以内的所有素数，并使用递归回溯算法来寻找所有可能的解决方案。

Prime Ring Problem

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

因为最大只到20，所以两个数字之和顶多也就39，那么可以直接把范围内的素数列出来然后用find函数就好。
剩下的还有能说的就是别忘了判断最后一个元素和1的和是不是素数。
关于空格和换行除了if-else也可以用条件运算符

代码

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
const vector<int> prime={2,3,5,7,11,13,17,19,23,29,31,37};
void circle(int pos,int n);
int isok(void);
void print(int n);
int arr[20],num[20];
int main()
{
    int t=0,x,i;
    while(cin>>x&&x>0&&x<20){
        cout<<"Case "<<++t<<":\n";
        for(i=0;i<x;++i){
            arr[i]=1;
            num[i]=0;
        }
        num[0]=1;
        circle(1,x);
        cout<<endl;
    }
}
void circle(int pos,int n)
{
    if(pos==n){
        if(find(prime.begin(),prime.end(),num[pos-1]+1)!=prime.end())//我把判断放在这里了
            print(n);
    }else{
        for(int i=1;i<n;++i){
            if(arr[i]&&find(prime.begin(),prime.end(),num[pos-1]+i+1)!=prime.end()){
                arr[i]=0;
                num[pos]=i+1;
                circle(pos+1,n);
                arr[i]=1;
            }
        }
    }
}
void print(int n)
{
    for(int i=0;i<n;++i){
        cout<<num[i];
        if(i==n-1){
            cout<<endl;
        }else{
            cout<<' ';
        }
    }
}