hdu 1016 Prime Ring Problem DFS解法 纪念我在杭电的第一百题

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29577    Accepted Submission(s): 13188

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
   
   
   
   

 我的第一百题和进杭电前一万名，，全贡献给了水题o(╯□╰)o

竟然格式错误两次！！

入门DFS。

下面代码：

#include <stdio.h>
#include <string.h>
#define MAX 25

bool visited[MAX] ;
int n , ans[MAX] ;
bool prime[2*MAX];

void DFS(int num)
{
	if(num == n)
	{
		if(!prime[ans[num-1]+1])
			return  ;
		for(int i = 0 ; i < n ; ++i)
		{
			printf("%d",ans[i]);
			if(i != n-1)
			{
				printf(" ");
			}
		}
		printf("\n") ;
		return ;
	}
	else
	{
		for(int i =  1 ; i <= n ; ++i)
		{
			if(!visited[i] && prime[i+ans[num-1]])
			{
				visited[i] = true ;
				ans[num] = i;
				DFS(num+1);
				visited[i] = false ;
			}
		}
	}
}

int main()
{
	for(int i = 2 ; i < 51 ; ++i)	prime[i] = true ;
	prime[0]=prime[1]=false;
	prime[2] = true ;
	for(int i = 2 ; i < 51; ++i)
	{
		for(int j = 2 ; j*i < 51 ; ++j)
		{
			prime[i*j] = false ;
		}
	}
	
	ans[0] = 1 ;
	int c=1;
	while(~scanf("%d",&n))
	{
		memset(visited,0,sizeof(visited));
		visited[1] = true ;
		printf("Case %d:\n",c++);
		DFS(1);
		printf("\n");
	}
	return 0 ;
}