HDU--杭电--1016--Prime Ring Problem--深度优先搜索--此类中的水题

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19807    Accepted Submission(s): 8858

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
#include
using namespace std;
int n,sushu[100]={0},huan[10]={0,1},see[22];  //see数组用来标记此数字是否可用，sushu数组是素数表，huan数组是记录素数环
void DFS(int s,int k)
{
    int i;
    if(s>n)
    {
        if(sushu[huan[1]+huan[s-1]])
        {
            for(i=1;i<=n;i++)
            {
                if(i>1)cout<<" ";
                cout<<huan[i];
            }cout<<endl;
        }
        else return;
    }
    for(i=2;i<=n;i++)
    {
        if(k)  //如果是第一次递归，那么把see数组初始化，不过我现在觉得没意义——！这个可以不要，删除没过的话找我是问
            for(int j=1;j<22;j++)
                see[j]=1;
        if(sushu[i+huan[s-1]]&&see[i])
        {
            huan[s]=i;//把i装进素数环的相应位置
            see[i]=0;
            DFS(s+1,0);  //指向下一个空洞-—-！并且用0标记以后的都不是第一次的递归
            see[i]=1;
        }
    }
}
int main (void)
{
    int i,j,k,l=1;
    for(i=1;i<100;i+=2)sushu[i]=1;  //虽然数量是比较少，不过我还是觉得打表爽一些
    sushu[1]=0;sushu[2]=1;
    for(i=3;i<100;i++)
        for(j=i+i;j<100;j+=i)
            sushu[j]=0;
        while(cin>>n)
        {
            cout<<"Case "<<l++<<":"<<endl;
            if(n==1)cout<<"1"<<endl;//经过无意间证明，其实不用这一句，而且要的话其实是错的，不信你贴我家代码输入1啊
            DFS(2,1);  //从环内第二个洞开始填，用1标记第一次的那个递归
            cout<<endl;
        }
    return 0;
}