HDUOJ----(1016)Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 21151    Accepted Submission(s): 9465

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)

深度搜索....无压力；；

代码：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int str[]={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20};
int ans[21]={1};
int n,cnt;  /*代表搜索的深度*/  
bool flag;
 /*可能需要剪枝*/
void  dfs(int step)
{
    int i,j,temp; 
   if(step==n)   /*说明搜索到底了！*/
   {
        flag=true;
         temp=ans[0]+ans[n-1];   //开头和结尾也要判断
           for(j=2;j*j<=temp;j++)
           {
            if(temp%j==0)
            {
               flag=false;
               break;
            }
           }
       if(flag)
       {
          printf("%d",ans[0]);
          for( i=1;i<n;i++)
          {
           printf(" %d",ans[i]);
          }
           puts("");
       }
   }
    else
    {
      for(i=1;i<n;i++)
      {
          if(str[i])
          {
             flag=true;     
             temp=ans[cnt-1]+str[i];
              for(j=2;j*j<=temp;j++)
              {
                  if(temp%j==0)
                  {
                     flag=false;
                      break;
                  }
              }
            if(flag)
            {
             ans[cnt++]=str[i];
             str[i]=0;
             dfs(step+1);
             str[i]=ans[--cnt];
             ans[cnt]=0;
            }
          }
      }
    }
}

int main()
{
    int count=1;
    while(scanf("%d",&n)!=EOF)
    {
        cnt=1;
        printf("Case %d:
",count++);
        dfs(1);
        puts("");
    }
    return 0;
}