HDU 1016 Prime Ring Problem

 

题意：给你一个N，要你把1->N这N个数组成一个环（第一个放的必须是1），要求是相邻的两个数加起来必须是素数（ 素数环 ），而且最后一个和第一个同样要保持这个性质

 

思路： 记忆化搜索

 

                                  Prime Ring Problem

                                             Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                             Total Submission(s): 18093    Accepted Submission(s): 8105

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
  
  

 

 

#include <stdio.h>
#include <algorithm>
#include <stdlib.h>
#include <string.h>
#include <math.h>
using namespace std;
int n,l;
int num[30];
bool visit[30]; 
bool flag;               //标记是否找到正解 

int judge( int x )
{
	for( int i=2 ; i<sqrt( (double)x )+1 ; i++ )
		if( x % i == 0 )
			return false;
	return true;
}

void DFS( int st , int count )
{
	visit[st] = true;
	if( count == n ){           //1到N全部放到了环里 
		if( judge( st + 1 ) ){
			//printf("enter\n");
			flag = true;   //搜到了结果 
		}
		return ;
	}
	for( int i=1 ; i<=n ; i++ ){
		if( !visit[i] && judge( st + i ) ){
			num[ count+1 ] = i;
			//printf("%d\n",i);
			DFS( i , count+1 );
			if( flag ){
				for( int j=1 ; j<=n ; j++ )
					printf( j==n ? "%d\n" : "%d " , num[j] );
				visit[i] = false;
				flag = false;
			}
			else
				visit[i] = false;
		}
	}
}	


int main( )
{
	int cas = 1;
	while( scanf("%d",&n) == 1 ){
		memset( visit , false , sizeof(visit) ); 
		flag = false;
		//l = 0;
		printf("Case %d:\n",cas++);
		num[1] = 1;
		DFS( 1 , 1 );
		printf("\n");
	}	
	return 0;
}