Hdu 1016 Prime Ring Problem (素数环经典dfs)

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20105    Accepted Submission(s): 9001

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)

Recommend

JGShining

 

题意：素数环 (经典搜索)

代码：

写法1：

#include
#include
#include
using namespace std;
int yes[22];
bool in[22];
bool prim[50];
int n;

void primtable()
{
    int i,j;
    memset(prim,0,sizeof(prim));
    for(i=3;i<=50;i++)
    {
        for(j=2;j<=i-1;j++)
        {
            if(i%j==0)
                prim[i]=true;
        }
    }
}

void dfs(int pos)
{
    int i,u;
    if(pos>n)
    {
        int m=yes[1]+yes[n];
        if(prim[m]==false)
        {
            for(i=1;i<=n;i++)
                printf("%d%c",yes[i],i==n?'\n':' ');
            //printf("\n");          //这里PE了一次
        }
        return;
    }
    for(i=2;i<=n;i++)
    {
		u=i+yes[pos-1];
        if(prim[u]==false&&in[i]==false)
        {
            yes[pos]=i;
            in[i]=true;
            dfs(pos+1);
            in[i]=false;
        }
    }
}

int main()
{
    int cas=1;
    primtable();
    while(scanf("%d",&n)!=EOF)
    {
        memset(in,0,sizeof(in));
        yes[1]=1;
        in[1]=true;
        printf("Case %d:\n",cas++);
        dfs(2);
        printf("\n");
    }
    return 0;
}

// 203MS

8765975

2013-07-30 16:19:31

Accepted

1016

203MS

228K

1125 B

C++

 

 

写法2：

#include
#include
#include
using namespace std;
int yes[22];
bool in[22];
bool prim[50];
int n;

void no_prim()
{
    int tmp,i,j;
    for(i=3;i<20;i+=2)
    {
        if(prim[i]==false)
        {
            tmp=i<<1;
            for(j=i*i;j<50;j+=tmp)
                prim[j]=true;
        }
    }
}

void dfs(int pos)
{
    int i,u;
    if(pos>n)
    {
        int m=yes[1]+yes[n];
        if(prim[m]==false&&m%2||m==2)
        {
            for(i=1;i<=n;i++)
                printf("%d%c",yes[i],i==n?'\n':' ');
            //printf("\n");
        }
        return;
    }
    for(i=2;i<=n;i++)
    {

        u=i+yes[pos-1];
        if((!prim[u]&&u%2||u==2)&&in[i]==false)
        {
            yes[pos]=i;
            in[i]=true;
            dfs(pos+1);
            in[i]=false;
        }
    }
}

int main()
{
    no_prim();
    int cas=1;
    while(scanf("%d",&n)!=EOF)
    {
        memset(in,0,sizeof(in));
        yes[1]=1;
        in[1]=true;
        printf("Case %d:\n",cas++);
        dfs(2);
        printf("\n");
    }
    return 0;
}

// 234MS

 

8765965

2013-07-30 16:18:52

Accepted

1016

234MS

228K

1143 B

C++