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2013
11-26

HDU 1016 Prime Ring Problem-DFS-[解题报告] C++

Prime Ring Problem

问题描述 :

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.

输入:

n (0 < n < 20).

输出:

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.

样例输入:

6
8

样例输出:

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:很容易理解,就是让你输出满足相邻的相加是素数的序列(注意不要重复)

思路就是深搜思想把每种情况遍历一次

代码实现:

#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];
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
   {
       for(i=0;i<num-1;i++)
           printf("%d ",a[i]);
       printf("%d\n",a[num-1]);
   }
   else
   {
       for(i=2;i<=n;i++)
       {
           if(visited[i]==0)//是否用过了
           {
               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));
         a[0]=1;
         dfs(1);
         printf("\n");
      }
      return 0;
}