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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本文探讨了PrimeRingProblem的解题策略，该问题要求在由n个圆组成的环上放置1至n的自然数，使得相邻两圆的数字之和为质数，且首个圆的数字必须为1。通过递归深度优先搜索算法，文章详细解释了如何找到所有可能的解决方案，并按字典序输出。

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 72721 Accepted Submission(s): 30979

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

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 20;

int vis[N],v[N],n,k=0;
bool Is_sushu(int a,int b)
{
    int s=a+b;
    for(int i=2;i<=sqrt(s);i++)
        if(s%i==0)
            return 0;
    return 1;
}

void dfs(int x)
{
    if(k==n&&Is_sushu(v[k-1],v[0]))
    {
        if(x>n) x=x%n;
        for(int i=0;i<n;i++)
        {
            if(i==n-1)
                printf("%d\n",v[i]);
            else
                printf("%d ",v[i]);
        }
    }
    else
    {
        for(int i=2;i<=n;i++)
        {
            if(Is_sushu(v[k-1],i)&&!vis[i])
            {
                v[k++]=i;
                vis[i]=1;
                dfs(i+1);
                k--;
                vis[i]=0;
            }
        }
    }
}

int main()
{
    int num=0;
    while(~scanf("%d",&n))
    {
        num++;
        printf("Case %d:\n",num);
        memset(vis,0,sizeof(vis));
        k=0,v[k++]=1,vis[0]=1;
        dfs(1);
        printf("\n");
    }
}