hdu1016 Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 41154    Accepted Submission(s): 18221

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

直接暴力搜索，我估计是需要剪枝，因为开始没有剪枝，然后测试了一下n=20，半天没有出结果。

于是想了一下剪枝：

1 开始的数字不是1的时候，说明所有的组合都找出来了，剩余的就是旋转这个圈得到的结果

2 如果当前的数字和它前一个相加不是素数的话，就不用继续往下找了。

 

#include<iostream>
#include<stdio.h>
using namespace std;
int n;
int num[25];
bool vis[25];
int prime[30];
void dfs(int a)
{
    if(a>n+1) return;
    if(num[1]!=1) return;
    if(a==n+1)
    {
        for(int i=1; i<n+1; i++)
        {
            if(!prime[num[i%n]+num[(i+1)%n]]) return;
        }
        //if(num[1]==1)
        // {
        for(int i=1; i<n; i++)
            printf("%d ",num[i%n]);
        printf("%d",num[0]);
        printf("
");
        // }
        return;
    }
    for(int i=1; i<=n; i++)
    {
        if(!vis[i])
        {
            num[a%n]=i;
            if(a>1)
            {
                if (!prime[num[a%n]+num[(a-1)%n]])
                continue;
            }
            //cout<<num[a%n]<<"  "<<endl;
            vis[i]=true;
            dfs(a+1);
            vis[i]=false;
        }

    }
}
int main()
{
    for(int i=0; i<100; i++) prime[i]=true;
    for(int i=2; i<50; i++)
        for(int j=i<<1; j<100; j+=i)
        {
            if(prime[j]) prime[j]=false;

        }
    int t=1;
    while((scanf("%d",&n))!=EOF)
    {
        for(int i=0; i<25; i++)
        {
            num[i]=0;
            vis[i]=false;
        }
        num[1]=1;
        printf("Case %d:
",t++);
        dfs(1);
        printf("
");
    }
    return 0;
}

 

Source

Asia 1996, Shanghai (Mainland China)

  

直接暴力搜索9