hdoj 1016 Prime Ring Problem【DFS】

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35069    Accepted Submission(s): 15523

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) 

题意：求出一个素数环，从1到n（输入的），相邻两个之和是素数并且第一个和最后一个的和也是素数，用深搜  递归处理

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
int n;
int a[25], use[25];
int isprime(int x)
{
	for(int i = 2; i < x/2; i++)
	{
		if(x%i==0)	return 0;
	}
	return 1;
}
void dfs(int c)
{
	int i,j;
	if(n==c&&isprime(1+a[n-1]))
	{
		for(i = 0; i < n; i++)
		printf(i!=n-1?"%d ":"%d\n",a[i]);
	}
	else
	{
		for(i = 2; i <= n; i++)
		{
			if(!use[i]&&isprime(i+a[c-1]))
			{
				a[c] = i;
				use[i] = 1;
				dfs(c+1);
				use[i] = 0;
			}
		}
	}
}
int main()
{
	int i,j,k,cas=1;
	while(scanf("%d",&n)==1)
	{
		memset(use, 0, sizeof(use));
		a[0] = 1;
		printf("Case %d:\n",cas++);
		dfs(1);
		printf("\n");
	}
	return 0;
}