Prime Ring Problem

Prime Ring Problem

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 13   Accepted Submission(s) : 5

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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)

这题我用广搜 递归没有理解

每次都从2开始 加 

 这一个组合最后一位+下一个没有被标记过的数 相加

并判断 这两个数的和是不是素数

如果是素数 就进队列；

如果 这个组合的最后一位 的位置与 n相同 就要判断他与头相加 是否为素数

#include<iostream>
#include<algorithm>
#include<queue>
struct word
{
    bool used[25];//记录该数字组合已使用过的数
    int end[25];//记录数字组合
    int ci;//记录次数
};
using namespace std;
int sushu[15]={1,2,3,5,7,11,13,17,19,23,29,31,37,41,43};//素数
//记录数据
bool check(int a)
{
    if(a==1)
        return 0;
    for(int i=0;i<15;i++)
    {
        if(a==sushu[i])
            return 1;//是素数
    
    }
    return 0;
}
int main()
{
    int n;
    int m=0;
    while(scanf("%d",&n)!=EOF)
    {
        m++;
        printf("Case %d:
",m);
        queue<word> q;
        word x1;
        x1.ci=1;
        memset(x1.used,0,sizeof(x1.used));
        memset(x1.end,0,sizeof(x1.end));
        x1.end[1]=1;
        x1.used[1]=1;//开头一位是1；
        q.push(x1);
        while(!q.empty())
        {
            word now;
            now=q.front();
            q.pop();
            if(now.ci==n)
            {
                if(check(now.end[n]+1))//到达最后一位 且 条件符合 输出
                {
                    printf("1");
                    for(int j=2;j<=n;j++)
                    {
                        printf(" %d",now.end[j]);
                    
                    }
                    printf("
");
                
                }
            
            }
            else 
            {
                for(int lin=2;lin<=n;lin++)
                {
                    if(now.used[lin]==0)
                    {
                        if(check(now.end[now.ci]+lin)&&now.ci<n)//如果是素数 现在的位置 加上下一位
                        {
                            now.used[lin]=1;
                            now.ci++;
                            now.end[now.ci]=lin;
                            q.push(now);//放进队列里
                            now.ci--;
                            now.used[lin]=0;//返回原值
                        }
                        if(check(now.end[now.ci]+now.end[1])&&now.ci==n)//若为最后一位 则与头相加；
                        {
                            now.used[lin]=1;
                            now.ci++;
                            now.end[now.ci]=lin;
                            q.push(now);
                            now.used[lin]=0;
                            now.ci--;

                            
                        
                        }
                    }


                }
            
            
            }
        
            
        }
        printf("
");
        

    }
            
    
        

    return 0;
}