HDU 1016 Prime Ring Problem （DFS回溯）

最新推荐文章于 2020-05-21 09:38:00 发布

yellow_hzy

最新推荐文章于 2020-05-21 09:38:00 发布

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本文探讨了素数环问题的解决方法，该问题要求在一个由n个节点组成的环中放置1到n的自然数，使得任意两个相邻节点上的数字之和为素数。文章提供了一段使用递归深度优先搜索算法的C++代码实现，用于寻找所有可能的环排列方案。

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Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 46987 Accepted Submission(s): 20747

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 <cstdio>
#include <cstdlib>
#include <iostream>
#include <set>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <queue>
using namespace std;
int primelist[30] = {0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1};
int a[29],vis[29],n;
void dfs(int x)
{
    if(x==n&&primelist[a[0]+a[n-1]]==1)//递归终止条件，1与a[n-1]在环上相邻
    {
        cout<<a[0];
        for(int i=1;i<n;i++)
            cout<<" "<<a[i];
        cout<<endl;
    }
    else
    {
        for(int i=2;i<=n;i++)//所有数都放置尝试
            if(!vis[i]&&primelist[i+a[x-1]])
        {
            a[x]=i;
            vis[i]=1;
            dfs(x+1);
            vis[i]=0;//清空标记
        }
    }
}
int main()
{
    a[0]=1;
    int ans=1;
    memset(vis,0,sizeof(vis));
    while(cin>>n)
    {
        printf("Case %d:\n",ans++);
        dfs(1);
        cout<<endl;
    }
    return 0;
}