HDU 1016 Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25182    Accepted Submission(s): 11243

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

68

 

Sample Output

Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2

#include <iostream>using namespace std;#define M 25int n,tot,su[M],vis[M],A[M];int S(int a){   int i;    for(i=2;i<=a/2;i++)    {        if(a%i==0)            return 0;    }    return 1;}//素数打表void DFS(int cur){   int i,j;        if(cur==n&&su[A[0]+A[n-1]])        //状态：cur(已经排列数的个数);范围：1<=x<=n;   解空间的边:n条,选择第i条边x=i;    {                                         for(i=0;i<n;i++)        {   if(i!=n-1)            printf("%d ",A[i]);        else             printf("%d",A[i]);        }        printf("\n");    }    else    {       for(i=2;i<=n;i++)       {                         if(su[A[cur-1]+i]&&!vis[i])  //如果两数相加不是素数或者这个数用过就直接跳过不做处理（处死）（限界函数：su[]和vis[]）           {                            //如果符合就表示可能存在解，进入子树，再将子树做新节点。               A[cur]=i;               vis[i]=1;               DFS(cur+1);               vis[i]=0;           }        }     }}int main(){   int i;    for(i=0;i<M;i++)    {        su[i]=S(i);        A[i]=i+1;    }    i=1;    while(scanf("%d",&n)!=EOF)    {         printf("Case %d:\n",i);        DFS(1);       i++;       printf("\n");    }    return 0;}/*解的构成：从根节点到叶节点的路径。限界函数：  (1)在从根节点开始到本节点的路径已经选过的不是合理选择。  (2)与父节点的和是素数。  (3)用过本节点的深度是n，与根节点的和为素数。如何处死：不处理。