HDU1016 素数环---(dfs)

http://acm.hdu.edu.cn/showproblem.php?pid=1016

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
题意:
输入整数n (0 < n < 20).
找出n个整数,能组成一个环,环中任意相邻两数之和为素数,字典序输出所有情况,每种情况1开头
#include<stdio.h>
#include<string.h>
int prime[40]={0,1,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0},n;//素数打表,因为n最大是20,所以只要打到40
int visited[21],a[21];//visited[]标记数组,a[]存符合条件的序列  定包含1->n   只是需要输出特定顺序的n个数//
void dfs(int num)//深搜
{
   int i;
   if(num==n&&prime[a[num-1]+a[0]])  //满足条件了,就输出来   然后一步步返回
   {
       for(i=0;i<num-1;i++)
           printf("%d ",a[i]);
       printf("%d\n",a[num-1]);
   }
   else                    //num  1->n-1
   {
       for(i=2;i<=n;i++)
       {
           if(visited[i]==0)//若未访问
           {
               if(prime[i+a[num-1]]) //是否和相邻的加起来是素数
               {
                   visited[i]=-1;//标记了
                   a[num++]=i;//放进数组
                   dfs(num); //递归调用
                   visited[i]=0; //退去标记
                   num--;   
               }
           }
       }
   }
}
int main()
{
      int num=0;
      while(scanf("%d",&n)!=EOF)
      {
         num++;
         printf("Case %d:\n",num);
         memset(visited,0,sizeof(visited));//标记数组初始化为0
         a[0]=1;
         dfs(1);
         printf("\n");
      }
      return 0;
}

 

posted @ 2017-03-11 11:24  kimsimple  阅读(193)  评论(0编辑  收藏  举报