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[]={,,,,,,,,,,,,,,,,,,,};
 int ans[]={};
 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[]+ans[n-];   //开头和结尾也要判断
            for(j=;j*j<=temp;j++)
            {
             if(temp%j==)
             {
                flag=false;
                break;
             }
            }
        if(flag)
        {
           printf("%d",ans[]);
           for( i=;i<n;i++)
           {
            printf(" %d",ans[i]);
           }
            puts("");
        }
    }
     else
     {
       for(i=;i<n;i++)
       {
           if(str[i])
           {
              flag=true;
              temp=ans[cnt-]+str[i];
               for(j=;j*j<=temp;j++)
               {
                   if(temp%j==)
                   {
                      flag=false;
                       break;
                   }
               }
             if(flag)
             {
              ans[cnt++]=str[i];
              str[i]=;
              dfs(step+);
              str[i]=ans[--cnt];
              ans[cnt]=;
             }
           }
       }
     }
 }

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