poj1016 Prime Ring Problem---dfs

最新推荐文章于 2014-03-30 17:17:20 发布

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本文提供了一个算法解决方案，用于生成一个由1到n的整数构成的环形排列，使得相邻两个数字之和为质数。通过递归深度优先搜索和质数检查，实现环形数字排列的生成。

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Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22718 Accepted Submission(s): 10116

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)

题目大意：给你一个n，输出用1~n的数组成一个环，相邻两数之间的和为素数。

代码：

#include <iostream>
#include <cstring>
using namespace std;

int hash[30];
int arr[100];
int n;

int check(int x)
{
    for(int i=2;i<x;i++)
    {
        if(x%i==0)
        return 0;
    }
    return 1;
}

void dfs(int order,int index)
{
    int i;
    arr[order]=index;
    hash[index]=1;

    if(order==n)
    {
        if(check(arr[order]+arr[1]))
        {
              cout<<"1";
              for(i=2;i<=n;i++)
              cout<<" "<<arr[i];
              cout<<endl;
        }
       return;
    }

    for(i=1;i<=n;i++)
    {
        if(!hash[i]&&check(arr[order]+i))
        {
            dfs(order+1,i);
            hash[i]=0;
        }
    }
    return;
}
int main()
{
    int flag=1;
    while(cin>>n)
    {
        cout<<"Case "<<flag++<<":"<<endl;

        memset(hash,0,sizeof(hash));
        dfs(1,1);

        cout<<endl;
    }
    return 0;
}